# Evaluate: (i) 3 √36 x 3 √384 (ii) 3 √96 x 3 √144 (iii) 3 √100 x 3 √270

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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) $\sqrt[3]{121}\times\sqrt[3]{297}$

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(i) $\sqrt[3]{36}\times\sqrt[3]{384}$

We have,

$\sqrt[3]{36}\times\sqrt[3]{384}$ = $\sqrt[3]{36\times384}$

∴ $(\sqrt[3]{a}$$\sqrt[3]{b}$ = $\sqrt[3]{ab})$

Now by prime factorization method,

$\sqrt[3]{36\times384}$

$2\times2\times2\times3 = 24.$

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

We have,

$\sqrt[3]{96}$ x $\sqrt[3]{122}$ = $\sqrt[3]{96\times122}$

$\because$ $(\sqrt[3]{a}$ x $\sqrt[3]{b}$ = $\sqrt[3]{ab}$)

Now by prime factorization method,

$\sqrt[3]{96\times122}$

$2\times2\times3 = 24.$

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

We have,

$\sqrt[3]{100}\times\sqrt[3]{270}$ = $\sqrt[3]{100\times270}$

$\because$   $(\sqrt[3]{a}$$\sqrt[3]{b}$ = $\sqrt[3]{ab})$

Now by prime factorization method,

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

$2\times3\times5 = 30.$

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

$\sqrt[3]{121}\times\sqrt[3]{297}$ = $\sqrt[3]{121\times297}$

$\because$    $(\sqrt[3]{a}$$\sqrt[3]{b}$ = $\sqrt[3]{ab})$

Now by prime factorization method,

$\sqrt[3]{121\times297}$ = $\sqrt[3]{{(11\times11)}\times(3\times\times3\times3\times11})$

$\sqrt[3]{11^3}$ x $\sqrt[3]{3^3}$

$11\times3 = 33.$