Divide (a^6+b^6)÷(a^2+b^2)

$(a^{6} + b^{6}) \div (a^{2} + b^{2})$

Step 1

Rewrite the division as a fraction.

$\frac{a^{6} + b^{6}}{a^{2} + b^{2}}$

Step 2

Simplify the numerator.

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

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

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

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

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 $b = b^{2}$ .

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

Simplify.

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

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

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

Multiply $2$ by $2$ .

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

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

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

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

Multiply $2$ by $2$ .

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

Step 3

Cancel the common factor of $a^{2} + b^{2}$ .

Cancel the common factor.

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

Divide $a^{4} - a^{2} b^{2} + b^{4}$ by $1$ .

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

Divide (a^6+b^6)÷(a^2+b^2)

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