Find the Product (x^2-2)^3

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

Step 1

Use the Binomial Theorem.

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

Step 2

Simplify each term.

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

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

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

Multiply $2$ by $3$ .

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

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

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

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

Multiply $2$ by $2$ .

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

Multiply $- 2$ by $3$ .

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

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

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

Multiply $4$ by $3$ .

$x^{6} - 6 x^{4} + 12 x^{2} + {(- 2)}^{3}$

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

$x^{6} - 6 x^{4} + 12 x^{2} - 8$