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

Answer 1

(i)
36 and 384 are not perfect cubes; therefore, we use the following property:
$\sqrt[3]{ab}=\sqrt[3]{a}\times \sqrt[3]{b}$ for any two integers a and b

$$\begin{gathered} \therefore \sqrt[3]{36}\times \sqrt[3]{384} \\ =\sqrt[3]{36\times 384} \end{gathered}$$

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

$$\begin{gathered} =\sqrt[3]{\left\{2\times 2\times 2\right\}\times \left\{2\times 2\times 2\right\}\times \left\{2\times 2\times 2\right\}\times \left\{3\times 3\times 3\right\}} \\ =2\times 2\times 2\times 3 \\ =24 \end{gathered}$$

Thus, the answer is 24.

(ii)
96 and 122 are not perfect cubes; therefore, we use the following property:
$\sqrt[3]{ab}=\sqrt[3]{a}\times \sqrt[3]{b}$ for any two integers a and b

$$\begin{gathered} \therefore \sqrt[3]{96}\times \sqrt[3]{144} \\ =\sqrt[3]{96\times 144} \end{gathered}$$

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

$$\begin{gathered} =\sqrt[3]{\left\{2\times 2\times 2\right\}\times \left\{2\times 2\times 2\right\}\times \left\{2\times 2\times 2\right\}\times \left\{3\times 3\times 3\right\}} \\ =2\times 2\times 2\times 3 \\ =24 \end{gathered}$$

Thus, the answer is 24.

(iii)
100 and 270 are not perfect cubes; therefore, we use the following property:
$\sqrt[3]{ab}=\sqrt[3]{a}\times \sqrt[3]{b}$ for any two integers a and b

$$\begin{gathered} \therefore \sqrt[3]{100}\times \sqrt[3]{270} \\ =\sqrt[3]{100\times 270} \end{gathered}$$

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

$$\begin{gathered} =\sqrt[3]{\left\{2\times 2\times 2\right\}\times \left\{3\times 3\times 3\right\}\times \left\{5\times 5\times 5\right\}} \\ =2\times 3\times 5 \\ =30 \end{gathered}$$

Thus, the answer is 30.

(iv)
121 and 297 are not perfect cubes; therefore, we use the following property:
$\sqrt[3]{ab}=\sqrt[3]{a}\times \sqrt[3]{b}$ for any two integers a and b

$$\begin{gathered} \therefore \sqrt[3]{121}\times \sqrt[3]{297} \\ =\sqrt[3]{121\times 297} \end{gathered}$$

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

$$\begin{gathered} =\sqrt[3]{\left\{11\times 11\times 11\right\}\times \left\{3\times 3\times 3\right\}} \\ =11\times 3 \\ =33 \end{gathered}$$

Thus, the answer is 33.