(i) \(\sqrt[3]{36}\times\sqrt[3]{384}\)
We have,
= \(\sqrt[3]{36}\times\sqrt[3]{384}\) = \(\sqrt[3]{36\times384}\)
∴ \((\sqrt[3]{a}\)x \(\sqrt[3]{b}\) = \(\sqrt[3]{ab})\)
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,
= \(\sqrt[3]{100}\times\sqrt[3]{270}\) = \(\sqrt[3]{100\times270}\)
\(\because\) \((\sqrt[3]{a}\)x \(\sqrt[3]{b}\) = \(\sqrt[3]{ab})\)
Now by prime factorization method,
= \(\sqrt[3]{100}\times\sqrt[3]{270}\)
=
=
=
=
= \(2\times3\times5 = 30.\)
(iv) \(\sqrt[3]{121}\times\sqrt[3]{297}\)
= \(\sqrt[3]{121}\times\sqrt[3]{297}\) = \(\sqrt[3]{121\times297}\)
\(\because\) \((\sqrt[3]{a}\)x \(\sqrt[3]{b}\) = \(\sqrt[3]{ab})\)
Now by prime factorization method,
= \(\sqrt[3]{121\times297}\) = \(\sqrt[3]{{(11\times11)}\times(3\times\times3\times3\times11})\)
= \(\sqrt[3]{11^3}\) x \(\sqrt[3]{3^3}\)
= \(11\times3 = 33.\)