Evaluate:

Question:

Evaluate:

(i) $\sqrt[3]{36} \times \sqrt[3]{384}$

(ii) $\sqrt[3]{96} \times \sqrt[3]{144}$

(iii) $\sqrt[3]{100} \times \sqrt[3]{270}$

(iv) $\sqrt[3]{121} \times \sqrt[3]{297}$

 

Solution:

(i)
36 and 384 are not perfect cubes; therefore, we use the following property:

$\sqrt[3]{a b}=\sqrt[3]{a} \times \sqrt[3]{b}$ for any two integers $a$ and $b$

$\therefore \sqrt[3]{36} \times \sqrt[3]{384}$

$=\sqrt[3]{36 \times 384}$

$=\sqrt[3]{(2 \times 2 \times 3 \times 3) \times(2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3)} \quad$ (By prime factorisation)

$=\sqrt[3]{\{2 \times 2 \times 2\} \times\{2 \times 2 \times 2\} \times\{2 \times 2 \times 2\} \times\{3 \times 3 \times 3\}}$

$=2 \times 2 \times 2 \times 3$

$=24$

Thus, the answer is 24.

(ii)
96 and 122 are not perfect cubes; therefore, we use the following property:

$\sqrt[3]{a b}=\sqrt[3]{a} \times \sqrt[3]{b}$ for any two integers $a$ and $b$

$\therefore \sqrt[3]{96} \times \sqrt[3]{144}$

$=\sqrt[3]{96} \times 144$

$=\sqrt[3]{(2 \times 2 \times 2 \times 2 \times 2 \times 3) \times(2 \times 2 \times 2 \times 2 \times 3 \times 3)}$     (By prime factorisation)

$=\sqrt[3]{\{2 \times 2 \times 2\} \times\{2 \times 2 \times 2\} \times\{2 \times 2 \times 2\} \times\{3 \times 3 \times 3\}}$

$=2 \times 2 \times 2 \times 3$

$=24$

Thus, the answer is 24.

(iii)
100 and 270 are not perfect cubes; therefore, we use the following property:

$\sqrt[3]{a b}=\sqrt[3]{a} \times \sqrt[3]{b}$ for any two integers $a$ and $b$

$\therefore \sqrt[3]{100} \times \sqrt[3]{270}$

$=\sqrt[3]{100 \times 270}$

$=\sqrt[3]{(2 \times 2 \times 5 \times 5) \times(2 \times 3 \times 3 \times 3 \times 5)}$ (By prime factorisation)

$=\sqrt[3]{\{2 \times 2 \times 2\} \times\{3 \times 3 \times 3\} \times\{5 \times 5 \times 5\}}$

$=2 \times 3 \times 5$

$=30$

Thus, the answer is 30.

(iv)
121 and 297 are not perfect cubes; therefore, we use the following property:

$\sqrt[3]{a b}=\sqrt[3]{a} \times \sqrt[3]{b}$ for any two integers $a$ and $b$

$\therefore \sqrt[3]{121} \times \sqrt[3]{297}$

$=\sqrt[3]{121 \times 297}$

$=\sqrt[3]{(11 \times 11) \times(3 \times 3 \times 3 \times 11)}$        (By prime factorisation)

$=\sqrt[3]{\{11 \times 11 \times 11\} \times\{3 \times 3 \times 3\}}$

$=11 \times 3$

$=33$

Thus, the answer is 33.

 

 

 

 

 

 

 

 

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