Expand Using the Binomial Theorem (x^2-1)^3

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

Step 1

Use the binomial expansion theorem to find each term. The binomial theorem states ${(a + b)}^{n} = \sum_{k = 0}^{n} n C k \cdot (a^{n - k} b^{k})$ .

$\sum_{k = 0}^{3} \frac{3!}{(3 - k)! k!} \cdot {(x^{2})}^{3 - k} \cdot {(- 1)}^{k}$

Step 2

Expand the summation.

$\frac{3!}{(3 - 0)! 0!} \cdot {(x^{2})}^{3 - 0} \cdot {(- 1)}^{0} + \frac{3!}{(3 - 1)! 1!} \cdot {(x^{2})}^{3 - 1} \cdot {(- 1)}^{1} + \frac{3!}{(3 - 2)! 2!} \cdot {(x^{2})}^{3 - 2} \cdot {(- 1)}^{2} + \frac{3!}{(3 - 3)! 3!} \cdot {(x^{2})}^{3 - 3} \cdot {(- 1)}^{3}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(x^{2})}^{3} \cdot {(- 1)}^{0} + 3 \cdot {(x^{2})}^{2} \cdot {(- 1)}^{1} + 3 \cdot {(x^{2})}^{1} \cdot {(- 1)}^{2} + 1 \cdot {(x^{2})}^{0} \cdot {(- 1)}^{3}$

Step 4

Simplify each term.

Multiply ${(x^{2})}^{3}$ by $1$ .

${(x^{2})}^{3} \cdot {(- 1)}^{0} + 3 \cdot {(x^{2})}^{2} \cdot {(- 1)}^{1} + 3 \cdot {(x^{2})}^{1} \cdot {(- 1)}^{2} + 1 \cdot {(x^{2})}^{0} \cdot {(- 1)}^{3}$

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

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

$x^{2 \cdot 3} \cdot {(- 1)}^{0} + 3 \cdot {(x^{2})}^{2} \cdot {(- 1)}^{1} + 3 \cdot {(x^{2})}^{1} \cdot {(- 1)}^{2} + 1 \cdot {(x^{2})}^{0} \cdot {(- 1)}^{3}$

Multiply $2$ by $3$ .

$x^{6} \cdot {(- 1)}^{0} + 3 \cdot {(x^{2})}^{2} \cdot {(- 1)}^{1} + 3 \cdot {(x^{2})}^{1} \cdot {(- 1)}^{2} + 1 \cdot {(x^{2})}^{0} \cdot {(- 1)}^{3}$

Anything raised to $0$ is $1$ .

$x^{6} \cdot 1 + 3 \cdot {(x^{2})}^{2} \cdot {(- 1)}^{1} + 3 \cdot {(x^{2})}^{1} \cdot {(- 1)}^{2} + 1 \cdot {(x^{2})}^{0} \cdot {(- 1)}^{3}$

Multiply $x^{6}$ by $1$ .

$x^{6} + 3 \cdot {(x^{2})}^{2} \cdot {(- 1)}^{1} + 3 \cdot {(x^{2})}^{1} \cdot {(- 1)}^{2} + 1 \cdot {(x^{2})}^{0} \cdot {(- 1)}^{3}$

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

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

$x^{6} + 3 \cdot x^{2 \cdot 2} \cdot {(- 1)}^{1} + 3 \cdot {(x^{2})}^{1} \cdot {(- 1)}^{2} + 1 \cdot {(x^{2})}^{0} \cdot {(- 1)}^{3}$

Multiply $2$ by $2$ .

$x^{6} + 3 \cdot x^{4} \cdot {(- 1)}^{1} + 3 \cdot {(x^{2})}^{1} \cdot {(- 1)}^{2} + 1 \cdot {(x^{2})}^{0} \cdot {(- 1)}^{3}$

Evaluate the exponent.

$x^{6} + 3 x^{4} \cdot - 1 + 3 \cdot {(x^{2})}^{1} \cdot {(- 1)}^{2} + 1 \cdot {(x^{2})}^{0} \cdot {(- 1)}^{3}$

Multiply $- 1$ by $3$ .

$x^{6} - 3 x^{4} + 3 \cdot {(x^{2})}^{1} \cdot {(- 1)}^{2} + 1 \cdot {(x^{2})}^{0} \cdot {(- 1)}^{3}$

Multiply the exponents in ${(x^{2})}^{1}$ .

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

$x^{6} - 3 x^{4} + 3 \cdot x^{2 \cdot 1} \cdot {(- 1)}^{2} + 1 \cdot {(x^{2})}^{0} \cdot {(- 1)}^{3}$

Multiply $2$ by $1$ .

$x^{6} - 3 x^{4} + 3 \cdot x^{2} \cdot {(- 1)}^{2} + 1 \cdot {(x^{2})}^{0} \cdot {(- 1)}^{3}$

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

$x^{6} - 3 x^{4} + 3 x^{2} \cdot 1 + 1 \cdot {(x^{2})}^{0} \cdot {(- 1)}^{3}$

Multiply $3$ by $1$ .

$x^{6} - 3 x^{4} + 3 x^{2} + 1 \cdot {(x^{2})}^{0} \cdot {(- 1)}^{3}$

Multiply ${(x^{2})}^{0}$ by $1$ .

$x^{6} - 3 x^{4} + 3 x^{2} + {(x^{2})}^{0} \cdot {(- 1)}^{3}$

Multiply the exponents in ${(x^{2})}^{0}$ .

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

$x^{6} - 3 x^{4} + 3 x^{2} + x^{2 \cdot 0} \cdot {(- 1)}^{3}$

Multiply $2$ by $0$ .

$x^{6} - 3 x^{4} + 3 x^{2} + x^{0} \cdot {(- 1)}^{3}$

Anything raised to $0$ is $1$ .

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

Multiply ${(- 1)}^{3}$ by $1$ .

$x^{6} - 3 x^{4} + 3 x^{2} + {(- 1)}^{3}$

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

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

Expand Using the Binomial Theorem (x^2-1)^3

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