Finding Cube root

**vaironxxrd** · October 16th 2011, 06:11 AM

Problem: $\sqrt[3]{\frac{8a^3}{27b^6}}$

Solution(Wrong):

$\frac{2a\sqrta}{3\sqrt{b \cdot b \cdot b \cdot b \cdot b \cdot b}}$

$\frac{2a\sqrta}{3 \cdot 3b^2}$

**e^(i*pi)** · October 16th 2011, 06:14 AM

Originally Posted by vaironxxrd

Problem: $\sqrt[3]{\frac{8a^3}{27b^6}}$

Solution(Wrong):

$\frac{2a\sqrta}{3\sqrt{b \cdot b \cdot b \cdot b \cdot b \cdot b}}$

$\frac{2a\sqrta}{3 \cdot 3b^2}$

$b^6 = (b^3)^2$

**skeeter** · October 16th 2011, 07:06 AM

looks like you have one too many "3's"

$\sqrt[3]{\frac{8a^3}{27b^6}} = \frac{\sqrt[3]{8} \cdot \sqrt[3]{a^3}}{\sqrt[3]{27} \cdot \sqrt[3]{b^6}}$

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