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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 \(\sqrt[3]{3^3}\)

\(11\times3 = 33.\)

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