Find the Product (1-a^2)^3

${(1 - a^{2})}^{3}$

Step 1

Use the Binomial Theorem.

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

Step 2

Simplify each term.

One to any power is one.

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

One to any power is one.

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

Multiply $3$ by $1$ .

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

Multiply $- 1$ by $3$ .

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

Multiply $3$ by $1$ .

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

Apply the product rule to $- a^{2}$ .

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

Raise $- 1$ to the power of $2$ .

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

Multiply ${(a^{2})}^{2}$ by $1$ .

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

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

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

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

Multiply $2$ by $2$ .

$1 - 3 a^{2} + 3 a^{4} + {(- a^{2})}^{3}$

Apply the product rule to $- a^{2}$ .

$1 - 3 a^{2} + 3 a^{4} + {(- 1)}^{3} {(a^{2})}^{3}$

Raise $- 1$ to the power of $3$ .

$1 - 3 a^{2} + 3 a^{4} - {(a^{2})}^{3}$

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

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

$1 - 3 a^{2} + 3 a^{4} - a^{2 \cdot 3}$

Multiply $2$ by $3$ .

$1 - 3 a^{2} + 3 a^{4} - a^{6}$