Q1. Find the cube roots of each of the following integers:
(i) -125
(ii) -5832
(iii) -2744000
(iv) -753571
(v) -32768
Sol.
$$ \text {We know that } \sqrt[3]{-x} = – \sqrt[3]{x} $$ $$ \text {(i) Cube root of -125 } = \sqrt[3]{-125} = -\sqrt[3]{125} $$ $$ = -\sqrt[3]{5 \times 5 \times 5} = -5 $$$$ \text {(ii) Cube root of -5832 } = \sqrt[3]{-5832} = -\sqrt[3]{5832} $$Using unit digits method, we have
The unit digit of 5832 is 2. Therefore, the unit digit in the cube root of 5832 will be 8.
After striking out the units, tens and hundreds of digits of 5832, we are left with 5.
Now, 1 is the largest number whose cube is less than or equal to 5.
Therefore, the tens digit of the cube root of 5832 is 1.
$$ \text {Hence, } \sqrt[3]{5832} = 18 $$ $$ \Rightarrow \quad \sqrt[3]{-5832} = -\sqrt[3]{5832} = -18 $$$$ \text {(iii) Cube root of -2744000 } = \sqrt[3]{-2744000} = -\sqrt[3]{2744000} $$ $$ =-\sqrt[3]{2744 \times 1000} $$ $$ =-\sqrt[3]{2744} \times \sqrt[3]{1000} $$$$ \sqrt[3]{1000} = \sqrt[3]{10 \times 10 \times 10} = 10 $$Lets find cube root of 2744, using unit digits method, so the unit digit of 2744 is 4.
Therefore, the unit digit in the cube root of 2744 will be 4.
After striking out the units, tens and hundreds of digits of 2744, we are left with 2.
Now, 1 is the largest number whose cube is less than or equal to 2.
Therefore, the tens digit of the cube root of 5832 is 1. $$ \text {Hence, } \sqrt[3]{2744} = 14 $$ $$ \Rightarrow \quad \sqrt[3]{-2744} = -\sqrt[3]{2744} = -14 $$ $$ \therefore \quad -\sqrt[3]{2744000} =-\sqrt[3]{2744} \times \sqrt[3]{1000} $$ $$ = -14 \times 10 = -140 $$$$ \text {(iv) Cube root of -753571 } = \sqrt[3]{-753571} = -\sqrt[3]{753571} $$Lets find cube root of 753571, using unit digits method, so the unit digit of 753571 is 1. Therefore, the unit digit in the cube root of 753571 will be 1.
After striking out the units, tens and hundreds of digits of 753571, we are left with 753.
$$ \because \quad 9^{3} \leq 753 \leq 10^{3} $$ $$ \therefore \text {The tens digit of the cube root of 753571 is 9.} $$ $$ \text {Hence, } \sqrt[3]{753571} = 91 $$ $$ \Rightarrow \quad \sqrt[3]{-753571} = -\sqrt[3]{753571} = -91 $$$$ \text {(v) Cube root of -32768 } = \sqrt[3]{-32768} = -\sqrt[3]{32768} $$Lets find cube root of 32768, using unit digits method, so the unit digit of 32768 is 8. Therefore, the unit digit in the cube root of 32768 will be 2.
After striking out the units, tens and hundreds of digits of 32768, we are left with 32.
$$ \because \quad 3^{3} \leq 32 \leq 4^{3} $$ $$ \therefore \text {The tens digit of the cube root of 32768 is 3.} $$ $$ \text {Hence, } \sqrt[3]{32768} = 32 $$ $$ \Rightarrow \quad \sqrt[3]{-32768} = -\sqrt[3]{32768} = -32 $$Q2. Show that:
\begin{array}{l} \text {(i) } \quad \sqrt[3]{27} \times \sqrt[3]{64}=\sqrt[3]{27 \times 64} \\ \text {(ii) } \quad \sqrt[3]{64 \times 729}=\sqrt[3]{64} \times \sqrt[3]{729} \\ \text {(iii) } \quad \sqrt[3]{-125 \times 216}=\sqrt[3]{-125} \times \sqrt[3]{216} \\ \text {(iv) } \quad \sqrt[3]{-125 x-1000}=\sqrt[3]{-125} \times \sqrt[3]{-1000} \\\text {Sol. } \\ \text {(i) } \quad \sqrt[3]{27} \times \sqrt[3]{64}=\sqrt[3]{27 \times 64} \\ LHS =\sqrt[3]{27} \times \sqrt[3]{64} \\ =\sqrt[3]{3 \times 3 \times 3} \times \sqrt[3]{4 \times 4 \times 4} \\ =3 \times 4 \\ =12 \\ RHS=\sqrt[3]{27 \times 64} \\ =\sqrt[3]{3 \times 3 \times 3 \times 4 \times 4 \times 4} \\ =3 \times 4 \\ =12 \\ \because \text {LHS = RHS, therefore, the given equation is verified. } \\ \\\text {(ii) } \quad \sqrt[3]{64 \times 729}=\sqrt[3]{64} \times \sqrt[3]{729} \\LHS =\sqrt[3]{64 \times 729} \\ =\sqrt[3]{4 \times 4 \times 4 \times 9 \times 9 \times 9} \\ =4 \times 9=36 \\RHS=\sqrt[3]{64} \times \sqrt[3]{729} \\ =\sqrt[3]{4 \times 4 \times 4} \times \sqrt[3]{9 \times 9 \times 9} \\ =4 \times 9=36 \\ \because \text {LHS = RHS, therefore, the given equation is verified. } \\ \\\text {(iii) } \quad \sqrt[3]{-125 \times 216}=\sqrt[3]{-125} \times \sqrt[3]{216} \\ LHS =\sqrt[3]{-125 \times 216} \\ =\sqrt[3]{-5 \times 5 \times 5 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3} \\ =-5 \times 2 \times 3=-30 \\RHS =\sqrt[3]{-125} \times \sqrt[3]{216} \\ =\sqrt[3]{-5 \times 5 \times 5} \times \sqrt[3]{2 \times 2 \times 2 \times 3 \times 3 \times 3} \\ =-5 \times 2 \times 3=-30 \\ \because \text {LHS = RHS, therefore, the given equation is verified. } \\ \\\text {(iv) } \quad \sqrt[3]{-125 \times -1000}=\sqrt[3]{-125} \times \sqrt[3]{-1000} \\LHS=\sqrt[3]{-125 x-1000} \\ =\sqrt[3]{-5 \times 5 \times 5 \times -10 \times 10 \times 10}=-5 \times -10=50 \\RHS = \sqrt[3]{-125} \times \sqrt[3]{-1000} \\ =\sqrt[3]{-5 \times 5 \times 5} \times \sqrt[3]{-10 \times 10 \times 10}=-5 \times -10 = 50 \\ \because \text {LHS = RHS, therefore, the given equation is verified. } \end{array}Q3. Find the cube root of each of the following numbers:
(i) 8 × 125
(ii) -1728 × 216
(iii) -27 × 2744
(iv) -729 × -15625$$ \text {Sol. We know that for any two integers x and y, } \sqrt[3]{(x \times y)}=\sqrt[3]{x} \times \sqrt[3]{y} $$(i) Cube root of 8 × 125
$$ \sqrt[3]{(8 \times 125}=\sqrt[3]{2^{3}} \times \sqrt[3]{5^{3}} $$ $$ =2 \times 5 =10 $$(ii) Cube root of -1728 × 216
$$ \sqrt[3]{(-1728 \times 216}= -\sqrt[3]{1728} \times \sqrt[3]{216} $$ $$ = -\sqrt[3]{12 \times 12 \times 12} \times \sqrt[3]{6 \times 6 \times 6} $$ $$ =-12 \times 6 =-72 $$(iii) Cube root of -27 × 2744
$$ \sqrt[3]{(-27 \times 2744}= -\sqrt[3]{27} \times \sqrt[3]{2744} $$ $$ = -\sqrt[3]{3 \times 3 \times 3} \times \sqrt[3]{14 \times 14 \times 14} $$ $$ =-3 \times 14 =-42 $$(iv) Cube root of -729 × -15625
$$ \sqrt[3]{(-729 \times -15625}= -\sqrt[3]{729} \times -\sqrt[3]{15625} $$ $$ = -\sqrt[3]{9 \times 9 \times 9} \times -\sqrt[3]{25 \times 25 \times 25} $$ $$ =-9 \times -25 = 225 $$Q4. Evaluate:
\begin{array}{l} \text {(i) } \quad \sqrt[3]{4^{3} \times 6^{3}} \quad \text {(ii) } \quad \sqrt[3]{8 \times 17 \times 17 \times 17} \\ \text {(iii) } \quad \sqrt[3]{700 \times 2 \times 49 \times 5} \quad \text {(iv) } \quad 125 \sqrt[3]{a^{6}}-\sqrt[3]{125 a^{6}} \\\text {Sol. We know that for any two integers x and y, } \sqrt[3]{(x \times y)}=\sqrt[3]{x} \times \sqrt[3]{y}\\\text {(i) } \quad \sqrt[3]{4^{3} \times 6^{3}} \\ \text {By using the above mentioned property, we have } \\ \sqrt[3]{(4^{3} \times 6^{3}}=\sqrt[3]{4^{3}} \times \sqrt[3]{6^{3}} \\ =4 \times 6 =24 \\ \\\text {(ii) } \quad \sqrt[3]{8 \times 17 \times 17 \times 17} \\ \text {By using the above mentioned property, we have } \\ \sqrt[3]{(8 \times 17 \times 17 \times 17)} =\sqrt[3]{8} \times \sqrt[3]{17 \times 17 \times 17} \\ =\sqrt[3]{2} \times \sqrt[3]{17^{3}} \\ =2 \times 17 =34 \\ \\\text {(iii) } \quad \sqrt[3]{700 \times 2 \times 49 \times 5} \\ =\sqrt[3]{(2 \times 2 \times 5 \times 5 \times 7 \times 2 \times 7 \times 7 \times 5)} \\ =\sqrt[3]{\left(2^{3} \times 5^{3} \times 7^{3}\right)} \\ =2 \times 5 \times 7 =70 \\ \\\text {(iv) } \quad 125 \sqrt[3]{a^{6}}-\sqrt[3]{125 a^{6}} \\ =125 \sqrt[3]{(a^{2})^{3}}-\sqrt[3] 5^{3}(a^{2})^{3} \\ =125 a^{2}- 5 a^{2} =120 a^{2}\end{array}Q5. Find the cube root of each of the following rational numbers:
\begin{array}{l} \text {(i) } \quad \frac{-125}{729} \quad \text {(ii) } \quad \frac{10648}{12167} \quad \text {(iii) } \quad \frac{-19683}{24389} \\ \text {(iv) } \quad \frac{686}{-3456} \quad \text {(v) } \quad \frac{-39304}{-42875} \\\text {Sol. We know that } \sqrt[3]{\frac {x}{y}} = \frac {\sqrt[3]{x}}{\sqrt[3]{y}} \text { and } \sqrt[3]{\frac {-x}{y}} = -\frac {\sqrt[3]{x}}{\sqrt[3]{y}} \\ \\\text {(i) } \quad \sqrt[3]{\frac {-125}{729}} =-\sqrt[3]{\frac {125}{729}} \\ =-\frac {\sqrt[3]{5 \times 5 \times 5}}{\sqrt[3]{9 \times 9 \times 9} }\\ = – \frac {5}{9} \quad \quad [\text {By taking one factor from each triple}]\\ \\\text {(ii) } \quad \sqrt[3]{\frac{10648}{12167}} \\=\frac {\sqrt[3]{2 \times 2 \times 2 \times 11 \times 11 \times 11}}{\sqrt[3]{23 \times 23 \times 23} }\\ = \frac {2 \times 11}{23} \quad \quad [\text {By taking one factor from each triple}] \\ = \frac {22}{23} \\ \\\text {(iii) } \quad \sqrt[3]{\frac{-19683}{24389}} \\ = – \sqrt[3]{\frac{19683}{24389}} \\ =-\frac {\sqrt[3]{3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3}}{\sqrt[3]{29 \times 29 \times 29} }\\ = -\frac {3 \times 3 \times 3}{29} \quad \quad [\text {By taking one factor from each triple}] \\ = -\frac {27}{29} \\ \\\text {(iv) } \quad \sqrt[3]{\frac{686}{-3456}} \\ = – \sqrt[3]{\frac{686}{3456}} \\ =-\sqrt[3] { \frac {2 \times 7 \times 7 \times 7}{2 \times 2^{6} \times 3^{3}} }\\ =-\sqrt[3] { \frac {7 \times 7 \times 7}{2^{6} \times 3^{3}} } \\=-\frac {\sqrt[3]{7 \times 7 \times 7}}{\sqrt[3]{2^{3} \times 2^{3} \times 3^{3}} }\\ = -\frac {7}{2 \times 2 \times 3} \quad \quad [\text {By taking one factor from each triple}] \\ = -\frac {7}{12} \\ \\\text {(v) } \quad \sqrt[3]{\frac{-39304}{-42875}} \\ = \sqrt[3]{\frac{39304}{42875}} \\ =\frac {\sqrt[3]{2 \times 2 \times 2 \times 17 \times 17 \times 17}}{\sqrt[3]{5 \times 5 \times 5 \times 7 \times 7 \times 7} }\\= \frac {2 \times 17}{5 \times 7} \quad \quad [\text {By taking one factor from each triple}] \\ = \frac {34}{35} \\ \\ \end{array}Q6. Find the cube root of each of the following rational numbers:
(i) 0.001728
(ii) 0.003375
(iii) 0.001
(iv) 1.331
\begin{array}{l} \text {Sol. We know that } \sqrt[3]{\frac {x}{y}} = \frac {\sqrt[3]{x}}{\sqrt[3]{y}} \text { and } \sqrt[3]{\frac {-x}{y}} = -\frac {\sqrt[3]{x}}{\sqrt[3]{y}} \\ \\\text {(i) } \quad \sqrt[3]{0.001728} =\sqrt[3]{\frac {1728}{1000000}} \\ =\frac {\sqrt[3]{1728}}{\sqrt[3]{1000000} }\\ =\frac {\sqrt[3]{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 }}{\sqrt[3]{100 \times 100 \times 100} }\\ = \frac {2 \times 2 \times 3}{100} \quad \quad [\text {By taking one factor from each triple}]\\ = \frac {12}{100} \\ = 0.12 \\ \\\text {(ii) } \quad \sqrt[3]{0.003375} =\sqrt[3]{\frac {3375}{1000000}} \\ =\frac {\sqrt[3]{3375}}{\sqrt[3]{1000000} }\\ =\frac {\sqrt[3]{3 \times 3 \times 3 \times 5 \times 5 \times 5 }}{\sqrt[3]{100 \times 100 \times 100} }\\ = \frac {3 \times 5}{100} \quad \quad [\text {By taking one factor from each triple}]\\ = \frac {15}{100} \\ = 0.15 \\ \\\text {(iii) } \quad \sqrt[3]{0.001} =\sqrt[3]{\frac {1}{1000}} \\ =\frac {\sqrt[3]{1 }}{\sqrt[3]{10 \times 10 \times 10} }\\ = \frac {1}{10} \quad \quad [\text {By taking one factor from each triple}]\\ = 0.1 \\ \\\text {(iv) } \quad \sqrt[3]{1.331} =\sqrt[3]{\frac {1331}{1000}} \\ =\frac {\sqrt[3]{3375}}{\sqrt[3]{1000} }\\ =\frac {\sqrt[3]{11 \times 11 \times 11 }}{\sqrt[3]{10 \times 10 \times 10} }\\ = \frac {11}{10} \quad \quad [\text {By taking one factor from each triple}]\\ = 1.1 \\ \\\end{array}Q7. Evaluate each of the following:
\begin{array}{l} \text {(i) } \quad \sqrt[3]{27}+\sqrt[3]{0.008}+\sqrt[3]{0.064} \\ \text {(ii) } \quad \sqrt[3]{1000}+\sqrt[3]{0.008}-\sqrt[3]{0.125} \quad \text {(iii) } \quad \sqrt[3]{\frac{729}{216}} \times \frac{6}{9} \\ \text {(iv) } \quad \sqrt[3]{\frac{0.027}{0.008}} \div \sqrt{\frac{0.09}{0.04}}-1 \\ \text {(v) } \quad \sqrt[3]{0.1 \times 0.1 \times 0.1 \times 13 \times 13 \times 13} \\\text {Sol. We know that } \sqrt[3]{\frac {x}{y}} = \frac {\sqrt[3]{x}}{\sqrt[3]{y}} \text { and } \sqrt[3]{\frac {-x}{y}} = -\frac {\sqrt[3]{x}}{\sqrt[3]{y}} \\ \\\text {(i) } \quad \sqrt[3]{27}+\sqrt[3]{0.008}+\sqrt[3]{0.064} \\=\sqrt[3]{3 \times 3 \times 3}+\sqrt[3]{\frac{8}{1000}}+\sqrt[3]{\frac{64}{1000}} \\ =\sqrt[3]{3 \times 3 \times 3}+\frac{\sqrt[3]{8}}{\sqrt[3]{1000}}+\frac{\sqrt[3]{64}}{\sqrt[3]{1000}} \\ =\sqrt[3]{3 \times 3 \times 3}+\frac{\sqrt[3]{2 \times 2 \times 2}}{\sqrt[3]{10 \times 10 \times 10}}+\frac{\sqrt[3]{4 \times 4 \times 4}}{\sqrt[3]{10 \times 10 \times 10}} \\ =3+\frac{2}{10}+\frac{4}{10} \\ =3+0.2+0.4=3.6 \\ \\\text {(ii) } \quad \sqrt[3]{1000}+\sqrt[3]{0.008}-\sqrt[3]{0.125} \\ =\sqrt[3]{10 \times 10 \times 10}+\sqrt[3]{\frac{8}{1000}}-\sqrt[3]{\frac{125}{1000}} \\ =\sqrt[3]{10 \times 10 \times 10}+\frac{\sqrt[3]{2 \times 2 \times 2}}{\sqrt[3]{10 \times 10 \times 10}}-\frac{\sqrt[3]{5 \times 5 \times 5}}{\sqrt[3]{10 \times 10 \times 10}} \\ =10+\frac{2}{10}-\frac{5}{10} \\ =10 + 0.2 – 0.5 = 9.7 \\ \\\text {(iii) } \quad \sqrt[3]{\frac{729}{216}} \times \frac{6}{9} \\ = \sqrt[3]{\frac{9 \times 9 \times 9}{2 \times 2 \times 2 \times 3 \times 3 \times 3}} \times \frac{6}{9} \\ =\frac{9}{2 \times 3} \times \frac{6}{9}=1 \\ \\\text {(iv) } \quad \sqrt[3]{\frac{0.027}{0.008}} \div \sqrt{\frac{0.09}{0.04}} – 1 \\ =\sqrt[3]{\frac{27}{8}} \div \sqrt{\frac{9}{4}} – 1 \\ =\sqrt[3]{\frac{3 \times 3 \times 3 }{2 \times 2 \times 2}} \div \sqrt{\frac{3 \times 3}{2 \times 2 }} – 1 \\= \frac{3}{2} \div \frac{3}{2} – 1 \quad \quad [\because \sqrt[3]{27} = 3, \sqrt{9} = 3]\\ = \frac{3}{2} \times \frac{2}{3} – 1 \\ = 1-1=0 \\ \\\text {(v) } \quad \sqrt[3]{0.1 \times 0.1 \times 0.1 \times 13 \times 13 \times 13} \\ = \sqrt[3]{\frac{1}{10} \times \frac{1}{10} \times \frac{1}{10} \times 13 \times 13 \times 13} \\ = \frac{13}{10} = 1.3 \\\end{array}
(i) -125
(ii) -5832
(iii) -2744000
(iv) -753571
(v) -32768
Sol.
$$ \text {We know that } \sqrt[3]{-x} = – \sqrt[3]{x} $$ $$ \text {(i) Cube root of -125 } = \sqrt[3]{-125} = -\sqrt[3]{125} $$ $$ = -\sqrt[3]{5 \times 5 \times 5} = -5 $$$$ \text {(ii) Cube root of -5832 } = \sqrt[3]{-5832} = -\sqrt[3]{5832} $$Using unit digits method, we have
The unit digit of 5832 is 2. Therefore, the unit digit in the cube root of 5832 will be 8.
After striking out the units, tens and hundreds of digits of 5832, we are left with 5.
Now, 1 is the largest number whose cube is less than or equal to 5.
Therefore, the tens digit of the cube root of 5832 is 1.
$$ \text {Hence, } \sqrt[3]{5832} = 18 $$ $$ \Rightarrow \quad \sqrt[3]{-5832} = -\sqrt[3]{5832} = -18 $$$$ \text {(iii) Cube root of -2744000 } = \sqrt[3]{-2744000} = -\sqrt[3]{2744000} $$ $$ =-\sqrt[3]{2744 \times 1000} $$ $$ =-\sqrt[3]{2744} \times \sqrt[3]{1000} $$$$ \sqrt[3]{1000} = \sqrt[3]{10 \times 10 \times 10} = 10 $$Lets find cube root of 2744, using unit digits method, so the unit digit of 2744 is 4.
Therefore, the unit digit in the cube root of 2744 will be 4.
After striking out the units, tens and hundreds of digits of 2744, we are left with 2.
Now, 1 is the largest number whose cube is less than or equal to 2.
Therefore, the tens digit of the cube root of 5832 is 1. $$ \text {Hence, } \sqrt[3]{2744} = 14 $$ $$ \Rightarrow \quad \sqrt[3]{-2744} = -\sqrt[3]{2744} = -14 $$ $$ \therefore \quad -\sqrt[3]{2744000} =-\sqrt[3]{2744} \times \sqrt[3]{1000} $$ $$ = -14 \times 10 = -140 $$$$ \text {(iv) Cube root of -753571 } = \sqrt[3]{-753571} = -\sqrt[3]{753571} $$Lets find cube root of 753571, using unit digits method, so the unit digit of 753571 is 1. Therefore, the unit digit in the cube root of 753571 will be 1.
After striking out the units, tens and hundreds of digits of 753571, we are left with 753.
$$ \because \quad 9^{3} \leq 753 \leq 10^{3} $$ $$ \therefore \text {The tens digit of the cube root of 753571 is 9.} $$ $$ \text {Hence, } \sqrt[3]{753571} = 91 $$ $$ \Rightarrow \quad \sqrt[3]{-753571} = -\sqrt[3]{753571} = -91 $$$$ \text {(v) Cube root of -32768 } = \sqrt[3]{-32768} = -\sqrt[3]{32768} $$Lets find cube root of 32768, using unit digits method, so the unit digit of 32768 is 8. Therefore, the unit digit in the cube root of 32768 will be 2.
After striking out the units, tens and hundreds of digits of 32768, we are left with 32.
$$ \because \quad 3^{3} \leq 32 \leq 4^{3} $$ $$ \therefore \text {The tens digit of the cube root of 32768 is 3.} $$ $$ \text {Hence, } \sqrt[3]{32768} = 32 $$ $$ \Rightarrow \quad \sqrt[3]{-32768} = -\sqrt[3]{32768} = -32 $$Q2. Show that:
\begin{array}{l} \text {(i) } \quad \sqrt[3]{27} \times \sqrt[3]{64}=\sqrt[3]{27 \times 64} \\ \text {(ii) } \quad \sqrt[3]{64 \times 729}=\sqrt[3]{64} \times \sqrt[3]{729} \\ \text {(iii) } \quad \sqrt[3]{-125 \times 216}=\sqrt[3]{-125} \times \sqrt[3]{216} \\ \text {(iv) } \quad \sqrt[3]{-125 x-1000}=\sqrt[3]{-125} \times \sqrt[3]{-1000} \\\text {Sol. } \\ \text {(i) } \quad \sqrt[3]{27} \times \sqrt[3]{64}=\sqrt[3]{27 \times 64} \\ LHS =\sqrt[3]{27} \times \sqrt[3]{64} \\ =\sqrt[3]{3 \times 3 \times 3} \times \sqrt[3]{4 \times 4 \times 4} \\ =3 \times 4 \\ =12 \\ RHS=\sqrt[3]{27 \times 64} \\ =\sqrt[3]{3 \times 3 \times 3 \times 4 \times 4 \times 4} \\ =3 \times 4 \\ =12 \\ \because \text {LHS = RHS, therefore, the given equation is verified. } \\ \\\text {(ii) } \quad \sqrt[3]{64 \times 729}=\sqrt[3]{64} \times \sqrt[3]{729} \\LHS =\sqrt[3]{64 \times 729} \\ =\sqrt[3]{4 \times 4 \times 4 \times 9 \times 9 \times 9} \\ =4 \times 9=36 \\RHS=\sqrt[3]{64} \times \sqrt[3]{729} \\ =\sqrt[3]{4 \times 4 \times 4} \times \sqrt[3]{9 \times 9 \times 9} \\ =4 \times 9=36 \\ \because \text {LHS = RHS, therefore, the given equation is verified. } \\ \\\text {(iii) } \quad \sqrt[3]{-125 \times 216}=\sqrt[3]{-125} \times \sqrt[3]{216} \\ LHS =\sqrt[3]{-125 \times 216} \\ =\sqrt[3]{-5 \times 5 \times 5 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3} \\ =-5 \times 2 \times 3=-30 \\RHS =\sqrt[3]{-125} \times \sqrt[3]{216} \\ =\sqrt[3]{-5 \times 5 \times 5} \times \sqrt[3]{2 \times 2 \times 2 \times 3 \times 3 \times 3} \\ =-5 \times 2 \times 3=-30 \\ \because \text {LHS = RHS, therefore, the given equation is verified. } \\ \\\text {(iv) } \quad \sqrt[3]{-125 \times -1000}=\sqrt[3]{-125} \times \sqrt[3]{-1000} \\LHS=\sqrt[3]{-125 x-1000} \\ =\sqrt[3]{-5 \times 5 \times 5 \times -10 \times 10 \times 10}=-5 \times -10=50 \\RHS = \sqrt[3]{-125} \times \sqrt[3]{-1000} \\ =\sqrt[3]{-5 \times 5 \times 5} \times \sqrt[3]{-10 \times 10 \times 10}=-5 \times -10 = 50 \\ \because \text {LHS = RHS, therefore, the given equation is verified. } \end{array}Q3. Find the cube root of each of the following numbers:
(i) 8 × 125
(ii) -1728 × 216
(iii) -27 × 2744
(iv) -729 × -15625$$ \text {Sol. We know that for any two integers x and y, } \sqrt[3]{(x \times y)}=\sqrt[3]{x} \times \sqrt[3]{y} $$(i) Cube root of 8 × 125
$$ \sqrt[3]{(8 \times 125}=\sqrt[3]{2^{3}} \times \sqrt[3]{5^{3}} $$ $$ =2 \times 5 =10 $$(ii) Cube root of -1728 × 216
$$ \sqrt[3]{(-1728 \times 216}= -\sqrt[3]{1728} \times \sqrt[3]{216} $$ $$ = -\sqrt[3]{12 \times 12 \times 12} \times \sqrt[3]{6 \times 6 \times 6} $$ $$ =-12 \times 6 =-72 $$(iii) Cube root of -27 × 2744
$$ \sqrt[3]{(-27 \times 2744}= -\sqrt[3]{27} \times \sqrt[3]{2744} $$ $$ = -\sqrt[3]{3 \times 3 \times 3} \times \sqrt[3]{14 \times 14 \times 14} $$ $$ =-3 \times 14 =-42 $$(iv) Cube root of -729 × -15625
$$ \sqrt[3]{(-729 \times -15625}= -\sqrt[3]{729} \times -\sqrt[3]{15625} $$ $$ = -\sqrt[3]{9 \times 9 \times 9} \times -\sqrt[3]{25 \times 25 \times 25} $$ $$ =-9 \times -25 = 225 $$Q4. Evaluate:
\begin{array}{l} \text {(i) } \quad \sqrt[3]{4^{3} \times 6^{3}} \quad \text {(ii) } \quad \sqrt[3]{8 \times 17 \times 17 \times 17} \\ \text {(iii) } \quad \sqrt[3]{700 \times 2 \times 49 \times 5} \quad \text {(iv) } \quad 125 \sqrt[3]{a^{6}}-\sqrt[3]{125 a^{6}} \\\text {Sol. We know that for any two integers x and y, } \sqrt[3]{(x \times y)}=\sqrt[3]{x} \times \sqrt[3]{y}\\\text {(i) } \quad \sqrt[3]{4^{3} \times 6^{3}} \\ \text {By using the above mentioned property, we have } \\ \sqrt[3]{(4^{3} \times 6^{3}}=\sqrt[3]{4^{3}} \times \sqrt[3]{6^{3}} \\ =4 \times 6 =24 \\ \\\text {(ii) } \quad \sqrt[3]{8 \times 17 \times 17 \times 17} \\ \text {By using the above mentioned property, we have } \\ \sqrt[3]{(8 \times 17 \times 17 \times 17)} =\sqrt[3]{8} \times \sqrt[3]{17 \times 17 \times 17} \\ =\sqrt[3]{2} \times \sqrt[3]{17^{3}} \\ =2 \times 17 =34 \\ \\\text {(iii) } \quad \sqrt[3]{700 \times 2 \times 49 \times 5} \\ =\sqrt[3]{(2 \times 2 \times 5 \times 5 \times 7 \times 2 \times 7 \times 7 \times 5)} \\ =\sqrt[3]{\left(2^{3} \times 5^{3} \times 7^{3}\right)} \\ =2 \times 5 \times 7 =70 \\ \\\text {(iv) } \quad 125 \sqrt[3]{a^{6}}-\sqrt[3]{125 a^{6}} \\ =125 \sqrt[3]{(a^{2})^{3}}-\sqrt[3] 5^{3}(a^{2})^{3} \\ =125 a^{2}- 5 a^{2} =120 a^{2}\end{array}Q5. Find the cube root of each of the following rational numbers:
\begin{array}{l} \text {(i) } \quad \frac{-125}{729} \quad \text {(ii) } \quad \frac{10648}{12167} \quad \text {(iii) } \quad \frac{-19683}{24389} \\ \text {(iv) } \quad \frac{686}{-3456} \quad \text {(v) } \quad \frac{-39304}{-42875} \\\text {Sol. We know that } \sqrt[3]{\frac {x}{y}} = \frac {\sqrt[3]{x}}{\sqrt[3]{y}} \text { and } \sqrt[3]{\frac {-x}{y}} = -\frac {\sqrt[3]{x}}{\sqrt[3]{y}} \\ \\\text {(i) } \quad \sqrt[3]{\frac {-125}{729}} =-\sqrt[3]{\frac {125}{729}} \\ =-\frac {\sqrt[3]{5 \times 5 \times 5}}{\sqrt[3]{9 \times 9 \times 9} }\\ = – \frac {5}{9} \quad \quad [\text {By taking one factor from each triple}]\\ \\\text {(ii) } \quad \sqrt[3]{\frac{10648}{12167}} \\=\frac {\sqrt[3]{2 \times 2 \times 2 \times 11 \times 11 \times 11}}{\sqrt[3]{23 \times 23 \times 23} }\\ = \frac {2 \times 11}{23} \quad \quad [\text {By taking one factor from each triple}] \\ = \frac {22}{23} \\ \\\text {(iii) } \quad \sqrt[3]{\frac{-19683}{24389}} \\ = – \sqrt[3]{\frac{19683}{24389}} \\ =-\frac {\sqrt[3]{3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3}}{\sqrt[3]{29 \times 29 \times 29} }\\ = -\frac {3 \times 3 \times 3}{29} \quad \quad [\text {By taking one factor from each triple}] \\ = -\frac {27}{29} \\ \\\text {(iv) } \quad \sqrt[3]{\frac{686}{-3456}} \\ = – \sqrt[3]{\frac{686}{3456}} \\ =-\sqrt[3] { \frac {2 \times 7 \times 7 \times 7}{2 \times 2^{6} \times 3^{3}} }\\ =-\sqrt[3] { \frac {7 \times 7 \times 7}{2^{6} \times 3^{3}} } \\=-\frac {\sqrt[3]{7 \times 7 \times 7}}{\sqrt[3]{2^{3} \times 2^{3} \times 3^{3}} }\\ = -\frac {7}{2 \times 2 \times 3} \quad \quad [\text {By taking one factor from each triple}] \\ = -\frac {7}{12} \\ \\\text {(v) } \quad \sqrt[3]{\frac{-39304}{-42875}} \\ = \sqrt[3]{\frac{39304}{42875}} \\ =\frac {\sqrt[3]{2 \times 2 \times 2 \times 17 \times 17 \times 17}}{\sqrt[3]{5 \times 5 \times 5 \times 7 \times 7 \times 7} }\\= \frac {2 \times 17}{5 \times 7} \quad \quad [\text {By taking one factor from each triple}] \\ = \frac {34}{35} \\ \\ \end{array}Q6. Find the cube root of each of the following rational numbers:
(i) 0.001728
(ii) 0.003375
(iii) 0.001
(iv) 1.331
\begin{array}{l} \text {Sol. We know that } \sqrt[3]{\frac {x}{y}} = \frac {\sqrt[3]{x}}{\sqrt[3]{y}} \text { and } \sqrt[3]{\frac {-x}{y}} = -\frac {\sqrt[3]{x}}{\sqrt[3]{y}} \\ \\\text {(i) } \quad \sqrt[3]{0.001728} =\sqrt[3]{\frac {1728}{1000000}} \\ =\frac {\sqrt[3]{1728}}{\sqrt[3]{1000000} }\\ =\frac {\sqrt[3]{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 }}{\sqrt[3]{100 \times 100 \times 100} }\\ = \frac {2 \times 2 \times 3}{100} \quad \quad [\text {By taking one factor from each triple}]\\ = \frac {12}{100} \\ = 0.12 \\ \\\text {(ii) } \quad \sqrt[3]{0.003375} =\sqrt[3]{\frac {3375}{1000000}} \\ =\frac {\sqrt[3]{3375}}{\sqrt[3]{1000000} }\\ =\frac {\sqrt[3]{3 \times 3 \times 3 \times 5 \times 5 \times 5 }}{\sqrt[3]{100 \times 100 \times 100} }\\ = \frac {3 \times 5}{100} \quad \quad [\text {By taking one factor from each triple}]\\ = \frac {15}{100} \\ = 0.15 \\ \\\text {(iii) } \quad \sqrt[3]{0.001} =\sqrt[3]{\frac {1}{1000}} \\ =\frac {\sqrt[3]{1 }}{\sqrt[3]{10 \times 10 \times 10} }\\ = \frac {1}{10} \quad \quad [\text {By taking one factor from each triple}]\\ = 0.1 \\ \\\text {(iv) } \quad \sqrt[3]{1.331} =\sqrt[3]{\frac {1331}{1000}} \\ =\frac {\sqrt[3]{3375}}{\sqrt[3]{1000} }\\ =\frac {\sqrt[3]{11 \times 11 \times 11 }}{\sqrt[3]{10 \times 10 \times 10} }\\ = \frac {11}{10} \quad \quad [\text {By taking one factor from each triple}]\\ = 1.1 \\ \\\end{array}Q7. Evaluate each of the following:
\begin{array}{l} \text {(i) } \quad \sqrt[3]{27}+\sqrt[3]{0.008}+\sqrt[3]{0.064} \\ \text {(ii) } \quad \sqrt[3]{1000}+\sqrt[3]{0.008}-\sqrt[3]{0.125} \quad \text {(iii) } \quad \sqrt[3]{\frac{729}{216}} \times \frac{6}{9} \\ \text {(iv) } \quad \sqrt[3]{\frac{0.027}{0.008}} \div \sqrt{\frac{0.09}{0.04}}-1 \\ \text {(v) } \quad \sqrt[3]{0.1 \times 0.1 \times 0.1 \times 13 \times 13 \times 13} \\\text {Sol. We know that } \sqrt[3]{\frac {x}{y}} = \frac {\sqrt[3]{x}}{\sqrt[3]{y}} \text { and } \sqrt[3]{\frac {-x}{y}} = -\frac {\sqrt[3]{x}}{\sqrt[3]{y}} \\ \\\text {(i) } \quad \sqrt[3]{27}+\sqrt[3]{0.008}+\sqrt[3]{0.064} \\=\sqrt[3]{3 \times 3 \times 3}+\sqrt[3]{\frac{8}{1000}}+\sqrt[3]{\frac{64}{1000}} \\ =\sqrt[3]{3 \times 3 \times 3}+\frac{\sqrt[3]{8}}{\sqrt[3]{1000}}+\frac{\sqrt[3]{64}}{\sqrt[3]{1000}} \\ =\sqrt[3]{3 \times 3 \times 3}+\frac{\sqrt[3]{2 \times 2 \times 2}}{\sqrt[3]{10 \times 10 \times 10}}+\frac{\sqrt[3]{4 \times 4 \times 4}}{\sqrt[3]{10 \times 10 \times 10}} \\ =3+\frac{2}{10}+\frac{4}{10} \\ =3+0.2+0.4=3.6 \\ \\\text {(ii) } \quad \sqrt[3]{1000}+\sqrt[3]{0.008}-\sqrt[3]{0.125} \\ =\sqrt[3]{10 \times 10 \times 10}+\sqrt[3]{\frac{8}{1000}}-\sqrt[3]{\frac{125}{1000}} \\ =\sqrt[3]{10 \times 10 \times 10}+\frac{\sqrt[3]{2 \times 2 \times 2}}{\sqrt[3]{10 \times 10 \times 10}}-\frac{\sqrt[3]{5 \times 5 \times 5}}{\sqrt[3]{10 \times 10 \times 10}} \\ =10+\frac{2}{10}-\frac{5}{10} \\ =10 + 0.2 – 0.5 = 9.7 \\ \\\text {(iii) } \quad \sqrt[3]{\frac{729}{216}} \times \frac{6}{9} \\ = \sqrt[3]{\frac{9 \times 9 \times 9}{2 \times 2 \times 2 \times 3 \times 3 \times 3}} \times \frac{6}{9} \\ =\frac{9}{2 \times 3} \times \frac{6}{9}=1 \\ \\\text {(iv) } \quad \sqrt[3]{\frac{0.027}{0.008}} \div \sqrt{\frac{0.09}{0.04}} – 1 \\ =\sqrt[3]{\frac{27}{8}} \div \sqrt{\frac{9}{4}} – 1 \\ =\sqrt[3]{\frac{3 \times 3 \times 3 }{2 \times 2 \times 2}} \div \sqrt{\frac{3 \times 3}{2 \times 2 }} – 1 \\= \frac{3}{2} \div \frac{3}{2} – 1 \quad \quad [\because \sqrt[3]{27} = 3, \sqrt{9} = 3]\\ = \frac{3}{2} \times \frac{2}{3} – 1 \\ = 1-1=0 \\ \\\text {(v) } \quad \sqrt[3]{0.1 \times 0.1 \times 0.1 \times 13 \times 13 \times 13} \\ = \sqrt[3]{\frac{1}{10} \times \frac{1}{10} \times \frac{1}{10} \times 13 \times 13 \times 13} \\ = \frac{13}{10} = 1.3 \\\end{array}