Expand Using the Binomial Theorem (y-5x)^3

${(y - 5 x)}^{3}$

Step 1

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

$\sum_{k = 0}^{3} \frac{3!}{(3 - k)! k!} \cdot {(y)}^{3 - k} \cdot {(- 5 x)}^{k}$

Step 2

Expand the summation.

$\frac{3!}{(3 - 0)! 0!} {(y)}^{3 - 0} \cdot {(- 5 x)}^{0} + \frac{3!}{(3 - 1)! 1!} {(y)}^{3 - 1} \cdot {(- 5 x)}^{1} + \frac{3!}{(3 - 2)! 2!} {(y)}^{3 - 2} \cdot {(- 5 x)}^{2} + \frac{3!}{(3 - 3)! 3!} {(y)}^{3 - 3} \cdot {(- 5 x)}^{3}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(y)}^{3} \cdot {(- 5 x)}^{0} + 3 \cdot {(y)}^{2} \cdot {(- 5 x)}^{1} + 3 \cdot {(y)}^{1} \cdot {(- 5 x)}^{2} + 1 \cdot {(y)}^{0} \cdot {(- 5 x)}^{3}$

Step 4

Simplify each term.

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

${(y)}^{3} \cdot {(- 5 x)}^{0} + 3 \cdot {(y)}^{2} \cdot {(- 5 x)}^{1} + 3 \cdot {(y)}^{1} \cdot {(- 5 x)}^{2} + 1 \cdot {(y)}^{0} \cdot {(- 5 x)}^{3}$

Apply the product rule to $- 5 x$ .

$y^{3} \cdot ({(- 5)}^{0} x^{0}) + 3 \cdot {(y)}^{2} \cdot {(- 5 x)}^{1} + 3 \cdot {(y)}^{1} \cdot {(- 5 x)}^{2} + 1 \cdot {(y)}^{0} \cdot {(- 5 x)}^{3}$

Rewrite using the commutative property of multiplication.

${(- 5)}^{0} y^{3} x^{0} + 3 \cdot {(y)}^{2} \cdot {(- 5 x)}^{1} + 3 \cdot {(y)}^{1} \cdot {(- 5 x)}^{2} + 1 \cdot {(y)}^{0} \cdot {(- 5 x)}^{3}$

Anything raised to $0$ is $1$ .

$1 y^{3} x^{0} + 3 \cdot {(y)}^{2} \cdot {(- 5 x)}^{1} + 3 \cdot {(y)}^{1} \cdot {(- 5 x)}^{2} + 1 \cdot {(y)}^{0} \cdot {(- 5 x)}^{3}$

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

$y^{3} x^{0} + 3 \cdot {(y)}^{2} \cdot {(- 5 x)}^{1} + 3 \cdot {(y)}^{1} \cdot {(- 5 x)}^{2} + 1 \cdot {(y)}^{0} \cdot {(- 5 x)}^{3}$

Anything raised to $0$ is $1$ .

$y^{3} \cdot 1 + 3 \cdot {(y)}^{2} \cdot {(- 5 x)}^{1} + 3 \cdot {(y)}^{1} \cdot {(- 5 x)}^{2} + 1 \cdot {(y)}^{0} \cdot {(- 5 x)}^{3}$

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

$y^{3} + 3 \cdot {(y)}^{2} \cdot {(- 5 x)}^{1} + 3 \cdot {(y)}^{1} \cdot {(- 5 x)}^{2} + 1 \cdot {(y)}^{0} \cdot {(- 5 x)}^{3}$

Simplify.

$y^{3} + 3 y^{2} \cdot (- 5 x) + 3 \cdot {(y)}^{1} \cdot {(- 5 x)}^{2} + 1 \cdot {(y)}^{0} \cdot {(- 5 x)}^{3}$

Rewrite using the commutative property of multiplication.

$y^{3} + 3 \cdot - 5 y^{2} x + 3 \cdot {(y)}^{1} \cdot {(- 5 x)}^{2} + 1 \cdot {(y)}^{0} \cdot {(- 5 x)}^{3}$

Multiply $3$ by $- 5$ .

$y^{3} - 15 y^{2} x + 3 \cdot {(y)}^{1} \cdot {(- 5 x)}^{2} + 1 \cdot {(y)}^{0} \cdot {(- 5 x)}^{3}$

Simplify.

$y^{3} - 15 y^{2} x + 3 \cdot y \cdot {(- 5 x)}^{2} + 1 \cdot {(y)}^{0} \cdot {(- 5 x)}^{3}$

Apply the product rule to $- 5 x$ .

$y^{3} - 15 y^{2} x + 3 y \cdot ({(- 5)}^{2} x^{2}) + 1 \cdot {(y)}^{0} \cdot {(- 5 x)}^{3}$

Rewrite using the commutative property of multiplication.

$y^{3} - 15 y^{2} x + 3 \cdot {(- 5)}^{2} y x^{2} + 1 \cdot {(y)}^{0} \cdot {(- 5 x)}^{3}$

Raise $- 5$ to the power of $2$ .

$y^{3} - 15 y^{2} x + 3 \cdot 25 y x^{2} + 1 \cdot {(y)}^{0} \cdot {(- 5 x)}^{3}$

Multiply $3$ by $25$ .

$y^{3} - 15 y^{2} x + 75 y x^{2} + 1 \cdot {(y)}^{0} \cdot {(- 5 x)}^{3}$

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

$y^{3} - 15 y^{2} x + 75 y x^{2} + {(y)}^{0} \cdot {(- 5 x)}^{3}$

Anything raised to $0$ is $1$ .

$y^{3} - 15 y^{2} x + 75 y x^{2} + 1 \cdot {(- 5 x)}^{3}$

Multiply ${(- 5 x)}^{3}$ by $1$ .

$y^{3} - 15 y^{2} x + 75 y x^{2} + {(- 5 x)}^{3}$

Apply the product rule to $- 5 x$ .

$y^{3} - 15 y^{2} x + 75 y x^{2} + {(- 5)}^{3} x^{3}$

Raise $- 5$ to the power of $3$ .

$y^{3} - 15 y^{2} x + 75 y x^{2} - 125 x^{3}$

Expand Using the Binomial Theorem (y-5x)^3

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