Question
Evaluate:
$\sqrt[3]{96}\times\sqrt[3]{144}$
$\sqrt[3]{96}\times\sqrt[3]{144}$
✓
Answer
96 and $122$ are not perfect cubes; therefore, we use the following property:
$\sqrt[3]{\text{ab}}=\sqrt[3]{\text{a}}\times\sqrt[3]{\text{b}}$ for any two integers a and b.
$\therefore\sqrt[3]{96}\times\sqrt[3]{144}$
$=\sqrt[3]{{96}\times{144}}$
$=\sqrt[3]{(2\times2\times2\times2\times 2\times3)\times(2\times2\times2\times2\times2\times3)}$ (By prime factorisation)
$=\sqrt[3]{\{2\times2\times2\}\times\{2\times2\times2\}\times\{2\times2\times2\}\times\{3\times3\times3\}}$
$=2\times2\times2\times3$
$=24$
Thus, the answer is $24.$
$\sqrt[3]{\text{ab}}=\sqrt[3]{\text{a}}\times\sqrt[3]{\text{b}}$ for any two integers a and b.
$\therefore\sqrt[3]{96}\times\sqrt[3]{144}$
$=\sqrt[3]{{96}\times{144}}$
$=\sqrt[3]{(2\times2\times2\times2\times 2\times3)\times(2\times2\times2\times2\times2\times3)}$ (By prime factorisation)
$=\sqrt[3]{\{2\times2\times2\}\times\{2\times2\times2\}\times\{2\times2\times2\}\times\{3\times3\times3\}}$
$=2\times2\times2\times3$
$=24$
Thus, the answer is $24.$
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