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

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

Step 1

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

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

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

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

Step 4

Simplify each term.

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

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

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

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

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

Multiply $3$ by $2$ .

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

Anything raised to $0$ is $1$ .

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

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

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

Multiply the exponents in ${(x^{3})}^{1}$ .

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

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

Multiply $3$ by $1$ .

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

Evaluate the exponent.

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

Multiply $- 1$ by $2$ .

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

Multiply ${(x^{3})}^{0}$ by $1$ .

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

Multiply the exponents in ${(x^{3})}^{0}$ .

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

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

Multiply $3$ by $0$ .

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

Anything raised to $0$ is $1$ .

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

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

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

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

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

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

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