# Step-by-step Solution

## Solve $\frac{\left(x^2+4\right)^{\left(\frac{1}{3}\right)}-2}{x^3-2x^2-16x+32}$

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### Videos

$\frac{\sqrt[3]{x^2+4}-2}{-\left(4+x\right)\left(2+\sqrt{x}\right)\left(-2+\sqrt{x}\right)\left(x-2\right)}$

## Step-by-step explanation

Problem to solve:

$\:\left(\left(\left(x^2+4\right)^{\left(1/3\right)}\right)-2\right)/\:\left(x^3-2x^2-16x+32\right)$
1

Divide $1$ by $3$

$\frac{\sqrt[3]{x^2+4}-2}{x^3-2x^2-16x+32}$
2

We can factor the polynomial $x^3-2x^2-16x+32$ using synthetic division (Ruffini's rule). We search for a root in the factors of the constant term $32$ and we found that $2$ is a root of the polynomial

$2^3-22^2-16\cdot 2+32=0$

$\frac{\sqrt[3]{x^2+4}-2}{-\left(4+x\right)\left(2+\sqrt{x}\right)\left(-2+\sqrt{x}\right)\left(x-2\right)}$
$\:\left(\left(\left(x^2+4\right)^{\left(1/3\right)}\right)-2\right)/\:\left(x^3-2x^2-16x+32\right)$