Simplify (1+x^2)^3

${(1 + x^{2})}^{3}$

Step 1

Use the Binomial Theorem.

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

Step 2

Simplify each term.

One to any power is one.

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

One to any power is one.

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

Multiply $3$ by $1$ .

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

Multiply $3$ by $1$ .

$1 + 3 x^{2} + 3 {(x^{2})}^{2} + {(x^{2})}^{3}$

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

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

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

Multiply $2$ by $2$ .

$1 + 3 x^{2} + 3 x^{4} + {(x^{2})}^{3}$

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

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

$1 + 3 x^{2} + 3 x^{4} + x^{2 \cdot 3}$

Multiply $2$ by $3$ .

$1 + 3 x^{2} + 3 x^{4} + x^{6}$