Factor by Grouping a^8-a^2b^6

$a^{8} - a^{2} b^{6}$

Factor out the GCF of $a^{2}$ from $a^{8} - a^{2} b^{6}$ .

Factor out the GCF of $a^{2}$ from each term in the polynomial.

Factor out the GCF of $a^{2}$ from the expression $a^{8}$ .

$a^{2} (a^{6}) - a^{2} b^{6}$

Factor out the GCF of $a^{2}$ from the expression $- a^{2} b^{6}$ .

$a^{2} (a^{6}) + a^{2} (- b^{6})$

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

$a^{2} (a^{6} - b^{6})$

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

$a^{2} ({(a^{2})}^{3} - b^{6})$

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

$a^{2} ({(a^{2})}^{3} - {(b^{2})}^{3})$

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 $a = a^{2}$ and $b = b^{2}$ .

$a^{2} ((a^{2} - b^{2}) ({(a^{2})}^{2} + a^{2} b^{2} + {(b^{2})}^{2}))$

Simplify.

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

$a^{2} ((a + b) (a - b) ({(a^{2})}^{2} + a^{2} b^{2} + {(b^{2})}^{2}))$

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

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

$a^{2} ((a + b) (a - b) (a^{2 \cdot 2} + a^{2} b^{2} + {(b^{2})}^{2}))$

Multiply $2$ by $2$ .

$a^{2} ((a + b) (a - b) (a^{4} + a^{2} b^{2} + {(b^{2})}^{2}))$

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

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

$a^{2} ((a + b) (a - b) (a^{4} + a^{2} b^{2} + b^{2 \cdot 2}))$

Multiply $2$ by $2$ .

$a^{2} ((a + b) (a - b) (a^{4} + a^{2} b^{2} + b^{4}))$

Factor.

Rewrite $a^{4} + a^{2} b^{2} + b^{4}$ in a factored form.

Rewrite the middle term.

$a^{2} ((a + b) (a - b) (a^{4} + 2 a^{2} b^{2} - a^{2} b^{2} + b^{4}))$

Rearrange terms.

$a^{2} ((a + b) (a - b) (a^{4} + 2 a^{2} b^{2} + b^{4} - a^{2} b^{2}))$

Factor first three terms by perfect square rule.

$a^{2} ((a + b) (a - b) ({(a^{2} + b^{2})}^{2} - a^{2} b^{2}))$

Rewrite $a^{2} b^{2}$ as ${(a b)}^{2}$ .

$a^{2} ((a + b) (a - b) ({(a^{2} + b^{2})}^{2} - {(a b)}^{2}))$

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

$a^{2} ((a + b) (a - b) ((a^{2} + b^{2} + a b) (a^{2} + b^{2} - (a b))))$

Remove parentheses.

$a^{2} ((a + b) (a - b) ((a^{2} + b^{2} + a b) (a^{2} + b^{2} - a b)))$

Remove unnecessary parentheses.

$a^{2} ((a + b) (a - b) (a^{2} + b^{2} + a b) (a^{2} + b^{2} - a b))$

Factor by Grouping a^8-a^2b^6

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