Evaluate sixth root of 64(x^2-3)^18

$\sqrt[6]{64 {(x^{2} - 3)}^{18}}$

Step 1

Rewrite $64 {(x^{2} - 3)}^{18}$ as ${(2 {(x^{2} - 3)}^{3})}^{6}$ .

$\sqrt[6]{{(2 {(x^{2} - 3)}^{3})}^{6}}$

Step 2

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

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

Step 3

Use the Binomial Theorem.

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

Step 4

Simplify terms.

Simplify each term.

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

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

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

Multiply $2$ by $3$ .

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

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

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

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

Multiply $2$ by $2$ .

$2 (x^{6} + 3 x^{4} \cdot - 3 + 3 x^{2} {(- 3)}^{2} + {(- 3)}^{3})$

Multiply $- 3$ by $3$ .

$2 (x^{6} - 9 x^{4} + 3 x^{2} {(- 3)}^{2} + {(- 3)}^{3})$

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

$2 (x^{6} - 9 x^{4} + 3 x^{2} \cdot 9 + {(- 3)}^{3})$

Multiply $9$ by $3$ .

$2 (x^{6} - 9 x^{4} + 27 x^{2} + {(- 3)}^{3})$

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

$2 (x^{6} - 9 x^{4} + 27 x^{2} - 27)$

Apply the distributive property.

$2 x^{6} + 2 (- 9 x^{4}) + 2 (27 x^{2}) + 2 \cdot - 27$

Step 5

Simplify.

Multiply $- 9$ by $2$ .

$2 x^{6} - 18 x^{4} + 2 (27 x^{2}) + 2 \cdot - 27$

Multiply $27$ by $2$ .

$2 x^{6} - 18 x^{4} + 54 x^{2} + 2 \cdot - 27$

Multiply $2$ by $- 27$ .

$2 x^{6} - 18 x^{4} + 54 x^{2} - 54$