Expand Using the Binomial Theorem (2u-3v)^3

${(2 u - 3 v)}^{3}$

Step 1

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

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

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

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

Step 4

Simplify each term.

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

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

Apply the product rule to $2 u$ .

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

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

$8 u^{3} \cdot {(- 3 v)}^{0} + 3 \cdot {(2 u)}^{2} \cdot {(- 3 v)}^{1} + 3 \cdot {(2 u)}^{1} \cdot {(- 3 v)}^{2} + 1 \cdot {(2 u)}^{0} \cdot {(- 3 v)}^{3}$

Apply the product rule to $- 3 v$ .

$8 u^{3} \cdot ({(- 3)}^{0} v^{0}) + 3 \cdot {(2 u)}^{2} \cdot {(- 3 v)}^{1} + 3 \cdot {(2 u)}^{1} \cdot {(- 3 v)}^{2} + 1 \cdot {(2 u)}^{0} \cdot {(- 3 v)}^{3}$

Rewrite using the commutative property of multiplication.

$8 \cdot {(- 3)}^{0} u^{3} v^{0} + 3 \cdot {(2 u)}^{2} \cdot {(- 3 v)}^{1} + 3 \cdot {(2 u)}^{1} \cdot {(- 3 v)}^{2} + 1 \cdot {(2 u)}^{0} \cdot {(- 3 v)}^{3}$

Anything raised to $0$ is $1$ .

$8 \cdot 1 u^{3} v^{0} + 3 \cdot {(2 u)}^{2} \cdot {(- 3 v)}^{1} + 3 \cdot {(2 u)}^{1} \cdot {(- 3 v)}^{2} + 1 \cdot {(2 u)}^{0} \cdot {(- 3 v)}^{3}$

Multiply $8$ by $1$ .

$8 u^{3} v^{0} + 3 \cdot {(2 u)}^{2} \cdot {(- 3 v)}^{1} + 3 \cdot {(2 u)}^{1} \cdot {(- 3 v)}^{2} + 1 \cdot {(2 u)}^{0} \cdot {(- 3 v)}^{3}$

Anything raised to $0$ is $1$ .

$8 u^{3} \cdot 1 + 3 \cdot {(2 u)}^{2} \cdot {(- 3 v)}^{1} + 3 \cdot {(2 u)}^{1} \cdot {(- 3 v)}^{2} + 1 \cdot {(2 u)}^{0} \cdot {(- 3 v)}^{3}$

Multiply $8$ by $1$ .

$8 u^{3} + 3 \cdot {(2 u)}^{2} \cdot {(- 3 v)}^{1} + 3 \cdot {(2 u)}^{1} \cdot {(- 3 v)}^{2} + 1 \cdot {(2 u)}^{0} \cdot {(- 3 v)}^{3}$

Apply the product rule to $2 u$ .

$8 u^{3} + 3 \cdot (2^{2} u^{2}) \cdot {(- 3 v)}^{1} + 3 \cdot {(2 u)}^{1} \cdot {(- 3 v)}^{2} + 1 \cdot {(2 u)}^{0} \cdot {(- 3 v)}^{3}$

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

$8 u^{3} + 3 \cdot (4 u^{2}) \cdot {(- 3 v)}^{1} + 3 \cdot {(2 u)}^{1} \cdot {(- 3 v)}^{2} + 1 \cdot {(2 u)}^{0} \cdot {(- 3 v)}^{3}$

Multiply $4$ by $3$ .

$8 u^{3} + 12 u^{2} \cdot {(- 3 v)}^{1} + 3 \cdot {(2 u)}^{1} \cdot {(- 3 v)}^{2} + 1 \cdot {(2 u)}^{0} \cdot {(- 3 v)}^{3}$

Simplify.

$8 u^{3} + 12 u^{2} \cdot (- 3 v) + 3 \cdot {(2 u)}^{1} \cdot {(- 3 v)}^{2} + 1 \cdot {(2 u)}^{0} \cdot {(- 3 v)}^{3}$

Rewrite using the commutative property of multiplication.

$8 u^{3} + 12 \cdot - 3 u^{2} v + 3 \cdot {(2 u)}^{1} \cdot {(- 3 v)}^{2} + 1 \cdot {(2 u)}^{0} \cdot {(- 3 v)}^{3}$

Multiply $12$ by $- 3$ .

$8 u^{3} - 36 u^{2} v + 3 \cdot {(2 u)}^{1} \cdot {(- 3 v)}^{2} + 1 \cdot {(2 u)}^{0} \cdot {(- 3 v)}^{3}$

Simplify.

$8 u^{3} - 36 u^{2} v + 3 \cdot (2 u) \cdot {(- 3 v)}^{2} + 1 \cdot {(2 u)}^{0} \cdot {(- 3 v)}^{3}$

Multiply $2$ by $3$ .

$8 u^{3} - 36 u^{2} v + 6 u \cdot {(- 3 v)}^{2} + 1 \cdot {(2 u)}^{0} \cdot {(- 3 v)}^{3}$

Apply the product rule to $- 3 v$ .

$8 u^{3} - 36 u^{2} v + 6 u \cdot ({(- 3)}^{2} v^{2}) + 1 \cdot {(2 u)}^{0} \cdot {(- 3 v)}^{3}$

Rewrite using the commutative property of multiplication.

$8 u^{3} - 36 u^{2} v + 6 \cdot {(- 3)}^{2} u v^{2} + 1 \cdot {(2 u)}^{0} \cdot {(- 3 v)}^{3}$

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

$8 u^{3} - 36 u^{2} v + 6 \cdot 9 u v^{2} + 1 \cdot {(2 u)}^{0} \cdot {(- 3 v)}^{3}$

Multiply $6$ by $9$ .

$8 u^{3} - 36 u^{2} v + 54 u v^{2} + 1 \cdot {(2 u)}^{0} \cdot {(- 3 v)}^{3}$

Multiply ${(2 u)}^{0}$ by $1$ .

$8 u^{3} - 36 u^{2} v + 54 u v^{2} + {(2 u)}^{0} \cdot {(- 3 v)}^{3}$

Apply the product rule to $2 u$ .

$8 u^{3} - 36 u^{2} v + 54 u v^{2} + 2^{0} u^{0} \cdot {(- 3 v)}^{3}$

Anything raised to $0$ is $1$ .

$8 u^{3} - 36 u^{2} v + 54 u v^{2} + 1 u^{0} \cdot {(- 3 v)}^{3}$

Multiply $u^{0}$ by $1$ .

$8 u^{3} - 36 u^{2} v + 54 u v^{2} + u^{0} \cdot {(- 3 v)}^{3}$

Anything raised to $0$ is $1$ .

$8 u^{3} - 36 u^{2} v + 54 u v^{2} + 1 \cdot {(- 3 v)}^{3}$

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

$8 u^{3} - 36 u^{2} v + 54 u v^{2} + {(- 3 v)}^{3}$

Apply the product rule to $- 3 v$ .

$8 u^{3} - 36 u^{2} v + 54 u v^{2} + {(- 3)}^{3} v^{3}$

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

$8 u^{3} - 36 u^{2} v + 54 u v^{2} - 27 v^{3}$

Expand Using the Binomial Theorem (2u-3v)^3

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