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

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

Step 1

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

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

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

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

Step 4

Simplify each term.

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

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

Apply the product rule to $5 y$ .

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

Rewrite using the commutative property of multiplication.

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

Anything raised to $0$ is $1$ .

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

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

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

Anything raised to $0$ is $1$ .

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

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

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

Simplify.

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

Rewrite using the commutative property of multiplication.

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

Multiply $3$ by $5$ .

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

Simplify.

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

Apply the product rule to $5 y$ .

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

Rewrite using the commutative property of multiplication.

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

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

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

Multiply $3$ by $25$ .

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

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

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

Anything raised to $0$ is $1$ .

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

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

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

Apply the product rule to $5 y$ .

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

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

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

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

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