Simplify square root of 7x^9+7

$\sqrt{7 x^{9} + 7}$

Step 1

Simplify with factoring out.

Factor $7$ out of $7 x^{9} + 7$ .

Factor $7$ out of $7 x^{9}$ .

$\sqrt{7 (x^{9}) + 7}$

Factor $7$ out of $7$ .

$\sqrt{7 (x^{9}) + 7 (1)}$

Factor $7$ out of $7 (x^{9}) + 7 (1)$ .

$\sqrt{7 (x^{9} + 1)}$

Simplify the expression.

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

$\sqrt{7 ({(x^{3})}^{3} + 1)}$

Rewrite $1$ as $1^{3}$ .

$\sqrt{7 ({(x^{3})}^{3} + 1^{3})}$

Step 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 $a = x^{3}$ and $b = 1$ .

$\sqrt{7 ((x^{3} + 1) ({(x^{3})}^{2} - x^{3} \cdot 1 + 1^{2}))}$

Step 3

Simplify.

Rewrite $1$ as $1^{3}$ .

$\sqrt{7 ((x^{3} + 1^{3}) ({(x^{3})}^{2} - x^{3} \cdot 1 + 1^{2}))}$

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 $a = x$ and $b = 1$ .

$\sqrt{7 ((x + 1) (x^{2} - x \cdot 1 + 1^{2}) ({(x^{3})}^{2} - x^{3} \cdot 1 + 1^{2}))}$

Simplify.

Multiply $- 1$ by $1$ .

$\sqrt{7 ((x + 1) (x^{2} - x + 1^{2}) ({(x^{3})}^{2} - x^{3} \cdot 1 + 1^{2}))}$

One to any power is one.

$\sqrt{7 ((x + 1) (x^{2} - x + 1) ({(x^{3})}^{2} - x^{3} \cdot 1 + 1^{2}))}$

Multiply $- 1$ by $1$ .

$\sqrt{7 ((x + 1) (x^{2} - x + 1) ({(x^{3})}^{2} - x^{3} + 1^{2}))}$

$\sqrt{7 (x + 1) (x^{2} - x + 1) ({(x^{3})}^{2} - x^{3} + 1^{2})}$

Step 4

Simplify each term.

Multiply the exponents in ${(x^{3})}^{2}$ .

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

$\sqrt{7 (x + 1) (x^{2} - x + 1) (x^{3 \cdot 2} - x^{3} + 1^{2})}$

Multiply $3$ by $2$ .

$\sqrt{7 (x + 1) (x^{2} - x + 1) (x^{6} - x^{3} + 1^{2})}$

One to any power is one.

$\sqrt{7 (x + 1) (x^{2} - x + 1) (x^{6} - x^{3} + 1)}$

Simplify square root of 7x^9+7

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