Simplify (z^3+64)/(z^3-4z^2+16z)

$\frac{z^{3} + 64}{z^{3} - 4 z^{2} + 16 z}$

Step 1

Simplify the numerator.

Rewrite $64$ as $4^{3}$ .

$\frac{z^{3} + 4^{3}}{z^{3} - 4 z^{2} + 16 z}$

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = z$ and $b = 4$ .

$\frac{(z + 4) (z^{2} - z \cdot 4 + 4^{2})}{z^{3} - 4 z^{2} + 16 z}$

Simplify.

Multiply $4$ by $- 1$ .

$\frac{(z + 4) (z^{2} - 4 z + 4^{2})}{z^{3} - 4 z^{2} + 16 z}$

Raise $4$ to the power of $2$ .

$\frac{(z + 4) (z^{2} - 4 z + 16)}{z^{3} - 4 z^{2} + 16 z}$

Step 2

Simplify terms.

Factor $z$ out of $z^{3} - 4 z^{2} + 16 z$ .

Factor $z$ out of $z^{3}$ .

$\frac{(z + 4) (z^{2} - 4 z + 16)}{z \cdot z^{2} - 4 z^{2} + 16 z}$

Factor $z$ out of $- 4 z^{2}$ .

$\frac{(z + 4) (z^{2} - 4 z + 16)}{z \cdot z^{2} + z (- 4 z) + 16 z}$

Factor $z$ out of $16 z$ .

$\frac{(z + 4) (z^{2} - 4 z + 16)}{z \cdot z^{2} + z (- 4 z) + z \cdot 16}$

Factor $z$ out of $z \cdot z^{2} + z (- 4 z)$ .

$\frac{(z + 4) (z^{2} - 4 z + 16)}{z (z^{2} - 4 z) + z \cdot 16}$

Factor $z$ out of $z (z^{2} - 4 z) + z \cdot 16$ .

$\frac{(z + 4) (z^{2} - 4 z + 16)}{z (z^{2} - 4 z + 16)}$

Cancel the common factor of $z^{2} - 4 z + 16$ .

Cancel the common factor.

$\frac{(z + 4) (z^{2} - 4 z + 16)}{z (z^{2} - 4 z + 16)}$

Rewrite the expression.

$\frac{z + 4}{z}$

Simplify (z^3+64)/(z^3-4z^2+16z)

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