Simplify - square root of (2x+1)^6

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

Step 1

Rewrite ${(2 x + 1)}^{6}$ as ${({(2 x + 1)}^{3})}^{2}$ .

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

Step 2

Pull terms out from under the radical, assuming positive real numbers.

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

Step 3

Use the Binomial Theorem.

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

Step 4

Simplify terms.

Simplify each term.

Apply the product rule to $2 x$ .

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

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

$- (8 x^{3} + 3 {(2 x)}^{2} \cdot 1 + 3 (2 x) \cdot 1^{2} + 1^{3})$

Apply the product rule to $2 x$ .

$- (8 x^{3} + 3 (2^{2} x^{2}) \cdot 1 + 3 (2 x) \cdot 1^{2} + 1^{3})$

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

$- (8 x^{3} + 3 (4 x^{2}) \cdot 1 + 3 (2 x) \cdot 1^{2} + 1^{3})$

Multiply $4$ by $3$ .

$- (8 x^{3} + 12 x^{2} \cdot 1 + 3 (2 x) \cdot 1^{2} + 1^{3})$

Multiply $12$ by $1$ .

$- (8 x^{3} + 12 x^{2} + 3 (2 x) \cdot 1^{2} + 1^{3})$

Multiply $2$ by $3$ .

$- (8 x^{3} + 12 x^{2} + 6 x \cdot 1^{2} + 1^{3})$

One to any power is one.

$- (8 x^{3} + 12 x^{2} + 6 x \cdot 1 + 1^{3})$

Multiply $6$ by $1$ .

$- (8 x^{3} + 12 x^{2} + 6 x + 1^{3})$

One to any power is one.

$- (8 x^{3} + 12 x^{2} + 6 x + 1)$

Apply the distributive property.

$- (8 x^{3}) - (12 x^{2}) - (6 x) - 1 \cdot 1$

Step 5

Simplify.

Multiply $8$ by $- 1$ .

$- 8 x^{3} - (12 x^{2}) - (6 x) - 1 \cdot 1$

Multiply $12$ by $- 1$ .

$- 8 x^{3} - 12 x^{2} - (6 x) - 1 \cdot 1$

Multiply $6$ by $- 1$ .

$- 8 x^{3} - 12 x^{2} - 6 x - 1 \cdot 1$

Multiply $- 1$ by $1$ .

$- 8 x^{3} - 12 x^{2} - 6 x - 1$