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

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

Step 1

Use the Binomial Theorem.

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

Step 2

Simplify each term.

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

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

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

Multiply $2$ by $3$ .

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

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

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

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

Multiply $2$ by $2$ .

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

Multiply $- 3$ by $3$ .

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

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

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

Multiply $9$ by $3$ .

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

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

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