# How to solve for $y$ in $x = \frac{\sqrt[3]{9y-5}}{\sqrt[3]{11}}$

I am trying to solve the following for $y$ but am lost. I tried to multiply by $\sqrt[3]{121}/\sqrt[3]{121}$ but don't think that is how to do it.

$$x = \frac{\sqrt[3]{9y-5}}{\sqrt[3]{11}}$$

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Let's cube both sides of the following: $$x = \frac{\sqrt[\large 3]{9y-5}}{\sqrt[3]{11}} = \sqrt[\large 3]{\frac{9y - 5}{11}}$$

That gives us: $$x^3 = \frac{9y - 5}{11}$$

Can you take it from here and isolate $y$?

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Always clear and clean. +1 – Amzoti Apr 12 '13 at 0:29

Hint: Cube both sides and it's nearly over.

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$y$ only occurs in one place, so you just need to make it the subject.

$\sqrt[3]{11}$ is just a constant, but the $y$ is trapped inside a cube root, so we want to get rid of that. What is the 'opposite' of a cube root? If we do that, what are we left with?

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$$x = \frac{\sqrt[3]{9y-5}}{\sqrt[3]{11}}$$

Cube both sides.

$$x^3=\frac{9y-5}{11}$$

Multiply both sides by $11$.

$$11x^3=9y-5$$

Add $5$ to both sides.

$$11x^3+5=9y$$

Divide both sides by $9$.

$$y=\frac{11x^3+5}{9}$$

Very easy.

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I apologize. I justified it (to myself), by the fact that mine was the only answer that completely solved this (very very very very easy) question. I realize that it was wrong and I have undone it. Sorry again, and feel free to flag. – John Marty Apr 12 '13 at 2:11