Evaluate (x^2+1)^3

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

Step 1

Use the Binomial Theorem.

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

Step 2

Simplify each term.

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

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

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

Multiply $2$ by $3$ .

$x^{6} + 3 {(x^{2})}^{2} \cdot 1 + 3 x^{2} \cdot 1^{2} + 1^{3}$

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

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

$x^{6} + 3 x^{2 \cdot 2} \cdot 1 + 3 x^{2} \cdot 1^{2} + 1^{3}$

Multiply $2$ by $2$ .

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

Multiply $3$ by $1$ .

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

One to any power is one.

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

Multiply $3$ by $1$ .

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

One to any power is one.

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