Expand Using the Binomial Theorem (y-2)^3

${(y - 2)}^{3}$

Step 1

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

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

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

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

Step 4

Simplify each term.

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

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

Anything raised to $0$ is $1$ .

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

Multiply $y^{3}$ by $1$ .

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

Evaluate the exponent.

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

Multiply $- 2$ by $3$ .

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

Simplify.

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

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

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

Multiply $4$ by $3$ .

$y^{3} - 6 y^{2} + 12 y + 1 \cdot {(y)}^{0} \cdot {(- 2)}^{3}$

Multiply ${(y)}^{0}$ by $1$ .

$y^{3} - 6 y^{2} + 12 y + {(y)}^{0} \cdot {(- 2)}^{3}$

Anything raised to $0$ is $1$ .

$y^{3} - 6 y^{2} + 12 y + 1 \cdot {(- 2)}^{3}$

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

$y^{3} - 6 y^{2} + 12 y + {(- 2)}^{3}$

Raise $- 2$ to the power of $3$ .

$y^{3} - 6 y^{2} + 12 y - 8$

Expand Using the Binomial Theorem (y-2)^3

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