Expand Using the Binomial Theorem (x-3y)^4

${(x - 3 y)}^{4}$

Step 1

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

$\sum_{k = 0}^{4} \frac{4!}{(4 - k)! k!} \cdot {(x)}^{4 - k} \cdot {(- 3 y)}^{k}$

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

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

Step 4

Simplify each term.

Multiply ${(x)}^{4}$ by $1$ .

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

Apply the product rule to $- 3 y$ .

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

Rewrite using the commutative property of multiplication.

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

Anything raised to $0$ is $1$ .

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

Multiply $x^{4}$ by $1$ .

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

Anything raised to $0$ is $1$ .

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

Multiply $x^{4}$ by $1$ .

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

Simplify.

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

Rewrite using the commutative property of multiplication.

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

Multiply $4$ by $- 3$ .

$x^{4} - 12 x^{3} y + 6 \cdot {(x)}^{2} \cdot {(- 3 y)}^{2} + 4 \cdot {(x)}^{1} \cdot {(- 3 y)}^{3} + 1 \cdot {(x)}^{0} \cdot {(- 3 y)}^{4}$

Apply the product rule to $- 3 y$ .

$x^{4} - 12 x^{3} y + 6 x^{2} \cdot ({(- 3)}^{2} y^{2}) + 4 \cdot {(x)}^{1} \cdot {(- 3 y)}^{3} + 1 \cdot {(x)}^{0} \cdot {(- 3 y)}^{4}$

Rewrite using the commutative property of multiplication.

$x^{4} - 12 x^{3} y + 6 \cdot {(- 3)}^{2} x^{2} y^{2} + 4 \cdot {(x)}^{1} \cdot {(- 3 y)}^{3} + 1 \cdot {(x)}^{0} \cdot {(- 3 y)}^{4}$

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

$x^{4} - 12 x^{3} y + 6 \cdot 9 x^{2} y^{2} + 4 \cdot {(x)}^{1} \cdot {(- 3 y)}^{3} + 1 \cdot {(x)}^{0} \cdot {(- 3 y)}^{4}$

Multiply $6$ by $9$ .

$x^{4} - 12 x^{3} y + 54 x^{2} y^{2} + 4 \cdot {(x)}^{1} \cdot {(- 3 y)}^{3} + 1 \cdot {(x)}^{0} \cdot {(- 3 y)}^{4}$

Simplify.

$x^{4} - 12 x^{3} y + 54 x^{2} y^{2} + 4 \cdot x \cdot {(- 3 y)}^{3} + 1 \cdot {(x)}^{0} \cdot {(- 3 y)}^{4}$

Apply the product rule to $- 3 y$ .

$x^{4} - 12 x^{3} y + 54 x^{2} y^{2} + 4 x \cdot ({(- 3)}^{3} y^{3}) + 1 \cdot {(x)}^{0} \cdot {(- 3 y)}^{4}$

Rewrite using the commutative property of multiplication.

$x^{4} - 12 x^{3} y + 54 x^{2} y^{2} + 4 \cdot {(- 3)}^{3} x y^{3} + 1 \cdot {(x)}^{0} \cdot {(- 3 y)}^{4}$

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

$x^{4} - 12 x^{3} y + 54 x^{2} y^{2} + 4 \cdot - 27 x y^{3} + 1 \cdot {(x)}^{0} \cdot {(- 3 y)}^{4}$

Multiply $4$ by $- 27$ .

$x^{4} - 12 x^{3} y + 54 x^{2} y^{2} - 108 x y^{3} + 1 \cdot {(x)}^{0} \cdot {(- 3 y)}^{4}$

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

$x^{4} - 12 x^{3} y + 54 x^{2} y^{2} - 108 x y^{3} + {(x)}^{0} \cdot {(- 3 y)}^{4}$

Anything raised to $0$ is $1$ .

$x^{4} - 12 x^{3} y + 54 x^{2} y^{2} - 108 x y^{3} + 1 \cdot {(- 3 y)}^{4}$

Multiply ${(- 3 y)}^{4}$ by $1$ .

$x^{4} - 12 x^{3} y + 54 x^{2} y^{2} - 108 x y^{3} + {(- 3 y)}^{4}$

Apply the product rule to $- 3 y$ .

$x^{4} - 12 x^{3} y + 54 x^{2} y^{2} - 108 x y^{3} + {(- 3)}^{4} y^{4}$

Raise $- 3$ to the power of $4$ .

$x^{4} - 12 x^{3} y + 54 x^{2} y^{2} - 108 x y^{3} + 81 y^{4}$

Expand Using the Binomial Theorem (x-3y)^4

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