How do you simplify $\frac{{(2 x^{3})}^{3} z^{2}}{2 z^{4}}$ $[(2x^3)^3 z^2] / (2z^4)$ ?

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Answer 1 · 2016-04-25T20:38:54

You first get rid of the brackets. ${(2 x^{3})}^{3} = 2^{3} \cdot x^{3 \cdot 3}$ $(2x^3)^3=2^3*x^(3*3)$

$=(8*x^9*z^2)/(2*z^4)=(cancel2*4*x^9*cancelz^2)/(cancel2*cancelz^2*z^2)=(4x^9)/z^2$

How do you simplify [(2x^3)^3 z^2] / (2z^4)(2x3)3z22z4[(2x^3)^3 z^2] / (2z^4)?