Factor 125x^3+8y^3

$125 x^{3} + 8 y^{3}$

Step 1

Rewrite $125 x^{3}$ as ${(5 x)}^{3}$ .

${(5 x)}^{3} + 8 y^{3}$

Step 2

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

${(5 x)}^{3} + {(2 y)}^{3}$

Step 3

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 $a = 5 x$ and $b = 2 y$ .

$(5 x + 2 y) ({(5 x)}^{2} - (5 x) (2 y) + {(2 y)}^{2})$

Step 4

Simplify.

Apply the product rule to $5 x$ .

$(5 x + 2 y) (5^{2} x^{2} - (5 x) (2 y) + {(2 y)}^{2})$

Raise $5$ to the power of $2$ .

$(5 x + 2 y) (25 x^{2} - (5 x) (2 y) + {(2 y)}^{2})$

Multiply $5$ by $- 1$ .

$(5 x + 2 y) (25 x^{2} - 5 x (2 y) + {(2 y)}^{2})$

Rewrite using the commutative property of multiplication.

$(5 x + 2 y) (25 x^{2} - 5 \cdot 2 x y + {(2 y)}^{2})$

Multiply $- 5$ by $2$ .

$(5 x + 2 y) (25 x^{2} - 10 x y + {(2 y)}^{2})$

Apply the product rule to $2 y$ .

$(5 x + 2 y) (25 x^{2} - 10 x y + 2^{2} y^{2})$

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

$(5 x + 2 y) (25 x^{2} - 10 x y + 4 y^{2})$

Factor 125x^3+8y^3

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