Find the Product (a^2-2b)^3

${(a^{2} - 2 b)}^{3}$

Step 1

Use the Binomial Theorem.

${(a^{2})}^{3} + 3 {(a^{2})}^{2} (- 2 b) + 3 a^{2} {(- 2 b)}^{2} + {(- 2 b)}^{3}$

Step 2

Simplify each term.

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

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

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

Multiply $2$ by $3$ .

$a^{6} + 3 {(a^{2})}^{2} (- 2 b) + 3 a^{2} {(- 2 b)}^{2} + {(- 2 b)}^{3}$

Rewrite using the commutative property of multiplication.

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

Multiply $3$ by $- 2$ .

$a^{6} - 6 {(a^{2})}^{2} b + 3 a^{2} {(- 2 b)}^{2} + {(- 2 b)}^{3}$

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

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

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

Multiply $2$ by $2$ .

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

Apply the product rule to $- 2 b$ .

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

Rewrite using the commutative property of multiplication.

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

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

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

Multiply $3$ by $4$ .

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

Apply the product rule to $- 2 b$ .

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

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

$a^{6} - 6 a^{4} b + 12 a^{2} b^{2} - 8 b^{3}$