Factor by Grouping 54x^3+250y^3

$54 x^{3} + 250 y^{3}$

Step 1

Factor out the GCF of $2$ from $54 x^{3} + 250 y^{3}$ .

Factor out the GCF of $2$ from each term in the polynomial.

Factor out the GCF of $2$ from the expression $54 x^{3}$ .

$2 (27 x^{3}) + 250 y^{3}$

Factor out the GCF of $2$ from the expression $250 y^{3}$ .

$2 (27 x^{3}) + 2 (125 y^{3})$

Since all the terms share a common factor of $2$ , it can be factored out of each term.

$2 (27 x^{3} + 125 y^{3})$

Step 2

Rewrite $27 x^{3}$ as ${(3 x)}^{3}$ .

$2 ({(3 x)}^{3} + 125 y^{3})$

Step 3

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

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

Step 4

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 $a = 3 x$ and $b = 5 y$ .

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

Step 5

Simplify.

Apply the product rule to $3 x$ .

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

Raise $3$ to the power of $2$ .

$2 ((3 x + 5 y) (9 x^{2} - 3 x (5 y) + {(5 y)}^{2}))$

Multiply $3$ by $- 1$ .

$2 ((3 x + 5 y) (9 x^{2} - 3 x (5 y) + {(5 y)}^{2}))$

Rewrite using the commutative property of multiplication.

$2 ((3 x + 5 y) (9 x^{2} - 3 \cdot (5 x y) + {(5 y)}^{2}))$

Multiply $- 3$ by $5$ .

$2 ((3 x + 5 y) (9 x^{2} - 15 x y + {(5 y)}^{2}))$

Apply the product rule to $5 y$ .

$2 ((3 x + 5 y) (9 x^{2} - 15 x y + 5^{2} y^{2}))$

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

$2 ((3 x + 5 y) (9 x^{2} - 15 x y + 25 y^{2}))$

Factor by Grouping 54x^3+250y^3

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