Factor a^6+b^6

$a^{6} + b^{6}$

Step 1

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

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

Step 2

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

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

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

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

Step 4

Simplify.

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

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

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

Multiply $2$ by $2$ .

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

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

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

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

Multiply $2$ by $2$ .

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

Factor a^6+b^6

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