How do you simplify $\sqrt{8 y^{3}} - \sqrt{2 y^{3}}$ $sqrt(8y^3)-sqrt(2y^3)$ ?

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Answer 1 · 2015-11-22T19:45:14

$y \sqrt{2 y}$ $ysqrt(2y)$

$\sqrt{8 y^{3}} - \sqrt{2 y^{3}}$ $sqrt(8y^3)-sqrt(2y^3)$

Factor $\sqrt{y^{3}} = \sqrt{y^{2} \cdot y} = y \sqrt{y}$ $sqrt{y^3}=sqrt(y^2*y)=ysqrty$

$= y \sqrt{y} \cdot (\sqrt{8} - \sqrt{2})$ $=ysqrty*(sqrt(8)-sqrt{2})$

Simplify $\sqrt{8} = \sqrt{4 \cdot 2} = 2 \sqrt{2}$ $\sqrt{8}=sqrt(4*2)=2sqrt2$ :

$= y \sqrt{y} \cdot (2 \sqrt{2} - \sqrt{2})$ $=ysqrty*(2\sqrt{2}-sqrt{2})$

$= y \sqrt{y} \cdot \sqrt{2}$ $=ysqrty*sqrt{2}$

$= y \sqrt{2 y}$ $=ysqrt{2y$

How do you simplify sqrt(8y^3)-sqrt(2y^3)√8y3−√2y3sqrt(8y^3)-sqrt(2y^3)?