Factor 27yx^6+y^7

$27 y x^{6} + y^{7}$

Step 1

Factor $y$ out of $27 y x^{6} + y^{7}$ .

Factor $y$ out of $27 y x^{6}$ .

$y (27 x^{6}) + y^{7}$

Factor $y$ out of $y^{7}$ .

$y (27 x^{6}) + y \cdot y^{6}$

Factor $y$ out of $y (27 x^{6}) + y \cdot y^{6}$ .

$y (27 x^{6} + y^{6})$

Step 2

Rewrite $27 x^{6}$ as ${(3 x^{2})}^{3}$ .

$y ({(3 x^{2})}^{3} + y^{6})$

Step 3

Rewrite $y^{6}$ as ${(y^{2})}^{3}$ .

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

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 $a = 3 x^{2}$ and $b = y^{2}$ .

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

Step 5

Factor.

Simplify.

Apply the product rule to $3 x^{2}$ .

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

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

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

Multiply the exponents in ${(x^{2})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Multiply $2$ by $2$ .

$y ((3 x^{2} + y^{2}) (9 x^{4} - (3 x^{2}) y^{2} + {(y^{2})}^{2}))$

Multiply $3$ by $- 1$ .

$y ((3 x^{2} + y^{2}) (9 x^{4} - 3 x^{2} y^{2} + {(y^{2})}^{2}))$

Multiply the exponents in ${(y^{2})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$y ((3 x^{2} + y^{2}) (9 x^{4} - 3 x^{2} y^{2} + y^{2 \cdot 2}))$

Multiply $2$ by $2$ .

$y ((3 x^{2} + y^{2}) (9 x^{4} - 3 x^{2} y^{2} + y^{4}))$

Factor.

Rewrite $9 x^{4} - 3 x^{2} y^{2} + y^{4}$ in a factored form.

Rewrite the middle term.

$y ((3 x^{2} + y^{2}) (9 x^{4} + 2 (3 x^{2}) y^{2} - 9 x^{2} y^{2} + y^{4}))$

Rearrange terms.

$y ((3 x^{2} + y^{2}) (9 x^{4} + 2 (3 x^{2}) y^{2} + y^{4} - 9 x^{2} y^{2}))$

Factor first three terms by perfect square rule.

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

Rewrite $9 x^{2} y^{2}$ as ${(3 x y)}^{2}$ .

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

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

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

Simplify.

Multiply $3$ by $- 1$ .

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

Remove parentheses.

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

Remove unnecessary parentheses.

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

Remove unnecessary parentheses.

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

Factor 27yx^6+y^7

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