Simplify (8x^3-y^3)/(4x^2-y^2)

$\frac{8 x^{3} - y^{3}}{4 x^{2} - y^{2}}$

Simplify the numerator.

Rewrite $8 x^{3}$ as ${(2 x)}^{3}$ .

$\frac{{(2 x)}^{3} - y^{3}}{4 x^{2} - y^{2}}$

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

$\frac{(2 x - y) ({(2 x)}^{2} + 2 x y + y^{2})}{4 x^{2} - y^{2}}$

Simplify.

Apply the product rule to $2 x$ .

$\frac{(2 x - y) (2^{2} x^{2} + 2 x y + y^{2})}{4 x^{2} - y^{2}}$

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

$\frac{(2 x - y) (4 x^{2} + 2 x y + y^{2})}{4 x^{2} - y^{2}}$

Simplify the denominator.

Rewrite $4 x^{2}$ as ${(2 x)}^{2}$ .

$\frac{(2 x - y) (4 x^{2} + 2 x y + y^{2})}{{(2 x)}^{2} - y^{2}}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2 x$ and $b = y$ .

$\frac{(2 x - y) (4 x^{2} + 2 x y + y^{2})}{(2 x + y) (2 x - y)}$

Cancel the common factor of $2 x - y$ .

Cancel the common factor.

$\frac{(2 x - y) (4 x^{2} + 2 x y + y^{2})}{(2 x + y) (2 x - y)}$

Rewrite the expression.

$\frac{4 x^{2} + 2 x y + y^{2}}{2 x + y}$

Simplify (8x^3-y^3)/(4x^2-y^2)

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