waecmaths question:
Simplify $\sqrt[3]{27{{x}^{3}}{{y}^{9}}}$
Option A:
$9x{{y}^{3}}$
Option B:
$3x{{y}^{6}}$
Option C:
$3x{{y}^{3}}$
Option D:
$9{{y}^{3}}$
waecmaths solution:
$\sqrt[3]{27{{x}^{3}}{{y}^{9}}}=\sqrt[3]{{{3}^{3}}{{x}^{3}}{{({{y}^{3}})}^{3}}}=3x{{y}^{3}}$
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