Question 10

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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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