Expand Using the Binomial Theorem (4a-3b)^4

${(4 a - 3 b)}^{4}$

Step 1

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

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

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

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

Step 4

Simplify each term.

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

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

Apply the product rule to $4 a$ .

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

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

$256 a^{4} \cdot {(- 3 b)}^{0} + 4 \cdot {(4 a)}^{3} \cdot {(- 3 b)}^{1} + 6 \cdot {(4 a)}^{2} \cdot {(- 3 b)}^{2} + 4 \cdot {(4 a)}^{1} \cdot {(- 3 b)}^{3} + 1 \cdot {(4 a)}^{0} \cdot {(- 3 b)}^{4}$

Apply the product rule to $- 3 b$ .

$256 a^{4} \cdot ({(- 3)}^{0} b^{0}) + 4 \cdot {(4 a)}^{3} \cdot {(- 3 b)}^{1} + 6 \cdot {(4 a)}^{2} \cdot {(- 3 b)}^{2} + 4 \cdot {(4 a)}^{1} \cdot {(- 3 b)}^{3} + 1 \cdot {(4 a)}^{0} \cdot {(- 3 b)}^{4}$

Rewrite using the commutative property of multiplication.

$256 \cdot {(- 3)}^{0} a^{4} b^{0} + 4 \cdot {(4 a)}^{3} \cdot {(- 3 b)}^{1} + 6 \cdot {(4 a)}^{2} \cdot {(- 3 b)}^{2} + 4 \cdot {(4 a)}^{1} \cdot {(- 3 b)}^{3} + 1 \cdot {(4 a)}^{0} \cdot {(- 3 b)}^{4}$

Anything raised to $0$ is $1$ .

$256 \cdot 1 a^{4} b^{0} + 4 \cdot {(4 a)}^{3} \cdot {(- 3 b)}^{1} + 6 \cdot {(4 a)}^{2} \cdot {(- 3 b)}^{2} + 4 \cdot {(4 a)}^{1} \cdot {(- 3 b)}^{3} + 1 \cdot {(4 a)}^{0} \cdot {(- 3 b)}^{4}$

Multiply $256$ by $1$ .

$256 a^{4} b^{0} + 4 \cdot {(4 a)}^{3} \cdot {(- 3 b)}^{1} + 6 \cdot {(4 a)}^{2} \cdot {(- 3 b)}^{2} + 4 \cdot {(4 a)}^{1} \cdot {(- 3 b)}^{3} + 1 \cdot {(4 a)}^{0} \cdot {(- 3 b)}^{4}$

Anything raised to $0$ is $1$ .

$256 a^{4} \cdot 1 + 4 \cdot {(4 a)}^{3} \cdot {(- 3 b)}^{1} + 6 \cdot {(4 a)}^{2} \cdot {(- 3 b)}^{2} + 4 \cdot {(4 a)}^{1} \cdot {(- 3 b)}^{3} + 1 \cdot {(4 a)}^{0} \cdot {(- 3 b)}^{4}$

Multiply $256$ by $1$ .

$256 a^{4} + 4 \cdot {(4 a)}^{3} \cdot {(- 3 b)}^{1} + 6 \cdot {(4 a)}^{2} \cdot {(- 3 b)}^{2} + 4 \cdot {(4 a)}^{1} \cdot {(- 3 b)}^{3} + 1 \cdot {(4 a)}^{0} \cdot {(- 3 b)}^{4}$

Apply the product rule to $4 a$ .

$256 a^{4} + 4 \cdot (4^{3} a^{3}) \cdot {(- 3 b)}^{1} + 6 \cdot {(4 a)}^{2} \cdot {(- 3 b)}^{2} + 4 \cdot {(4 a)}^{1} \cdot {(- 3 b)}^{3} + 1 \cdot {(4 a)}^{0} \cdot {(- 3 b)}^{4}$

Multiply $4$ by $4^{3}$ by adding the exponents.

Move $4^{3}$ .

$256 a^{4} + 4^{3} \cdot 4 \cdot a^{3} \cdot {(- 3 b)}^{1} + 6 \cdot {(4 a)}^{2} \cdot {(- 3 b)}^{2} + 4 \cdot {(4 a)}^{1} \cdot {(- 3 b)}^{3} + 1 \cdot {(4 a)}^{0} \cdot {(- 3 b)}^{4}$

Multiply $4^{3}$ by $4$ .

Raise $4$ to the power of $1$ .

$256 a^{4} + 4^{3} \cdot 4^{1} \cdot a^{3} \cdot {(- 3 b)}^{1} + 6 \cdot {(4 a)}^{2} \cdot {(- 3 b)}^{2} + 4 \cdot {(4 a)}^{1} \cdot {(- 3 b)}^{3} + 1 \cdot {(4 a)}^{0} \cdot {(- 3 b)}^{4}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$256 a^{4} + 4^{3 + 1} \cdot a^{3} \cdot {(- 3 b)}^{1} + 6 \cdot {(4 a)}^{2} \cdot {(- 3 b)}^{2} + 4 \cdot {(4 a)}^{1} \cdot {(- 3 b)}^{3} + 1 \cdot {(4 a)}^{0} \cdot {(- 3 b)}^{4}$

Add $3$ and $1$ .

$256 a^{4} + 4^{4} \cdot a^{3} \cdot {(- 3 b)}^{1} + 6 \cdot {(4 a)}^{2} \cdot {(- 3 b)}^{2} + 4 \cdot {(4 a)}^{1} \cdot {(- 3 b)}^{3} + 1 \cdot {(4 a)}^{0} \cdot {(- 3 b)}^{4}$

Simplify $4^{4} \cdot a^{3} \cdot {(- 3 b)}^{1}$ .

$256 a^{4} + 4^{4} \cdot a^{3} \cdot (- 3 b) + 6 \cdot {(4 a)}^{2} \cdot {(- 3 b)}^{2} + 4 \cdot {(4 a)}^{1} \cdot {(- 3 b)}^{3} + 1 \cdot {(4 a)}^{0} \cdot {(- 3 b)}^{4}$

Rewrite using the commutative property of multiplication.

$256 a^{4} + 4^{4} \cdot - 3 a^{3} b + 6 \cdot {(4 a)}^{2} \cdot {(- 3 b)}^{2} + 4 \cdot {(4 a)}^{1} \cdot {(- 3 b)}^{3} + 1 \cdot {(4 a)}^{0} \cdot {(- 3 b)}^{4}$

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

$256 a^{4} + 256 \cdot - 3 a^{3} b + 6 \cdot {(4 a)}^{2} \cdot {(- 3 b)}^{2} + 4 \cdot {(4 a)}^{1} \cdot {(- 3 b)}^{3} + 1 \cdot {(4 a)}^{0} \cdot {(- 3 b)}^{4}$

Multiply $256$ by $- 3$ .

$256 a^{4} - 768 a^{3} b + 6 \cdot {(4 a)}^{2} \cdot {(- 3 b)}^{2} + 4 \cdot {(4 a)}^{1} \cdot {(- 3 b)}^{3} + 1 \cdot {(4 a)}^{0} \cdot {(- 3 b)}^{4}$

Apply the product rule to $4 a$ .

$256 a^{4} - 768 a^{3} b + 6 \cdot (4^{2} a^{2}) \cdot {(- 3 b)}^{2} + 4 \cdot {(4 a)}^{1} \cdot {(- 3 b)}^{3} + 1 \cdot {(4 a)}^{0} \cdot {(- 3 b)}^{4}$

Raise $4$ to the power of $2$ .

$256 a^{4} - 768 a^{3} b + 6 \cdot (16 a^{2}) \cdot {(- 3 b)}^{2} + 4 \cdot {(4 a)}^{1} \cdot {(- 3 b)}^{3} + 1 \cdot {(4 a)}^{0} \cdot {(- 3 b)}^{4}$

Multiply $16$ by $6$ .

$256 a^{4} - 768 a^{3} b + 96 a^{2} \cdot {(- 3 b)}^{2} + 4 \cdot {(4 a)}^{1} \cdot {(- 3 b)}^{3} + 1 \cdot {(4 a)}^{0} \cdot {(- 3 b)}^{4}$

Apply the product rule to $- 3 b$ .

$256 a^{4} - 768 a^{3} b + 96 a^{2} \cdot ({(- 3)}^{2} b^{2}) + 4 \cdot {(4 a)}^{1} \cdot {(- 3 b)}^{3} + 1 \cdot {(4 a)}^{0} \cdot {(- 3 b)}^{4}$

Rewrite using the commutative property of multiplication.

$256 a^{4} - 768 a^{3} b + 96 \cdot {(- 3)}^{2} a^{2} b^{2} + 4 \cdot {(4 a)}^{1} \cdot {(- 3 b)}^{3} + 1 \cdot {(4 a)}^{0} \cdot {(- 3 b)}^{4}$

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

$256 a^{4} - 768 a^{3} b + 96 \cdot 9 a^{2} b^{2} + 4 \cdot {(4 a)}^{1} \cdot {(- 3 b)}^{3} + 1 \cdot {(4 a)}^{0} \cdot {(- 3 b)}^{4}$

Multiply $96$ by $9$ .

$256 a^{4} - 768 a^{3} b + 864 a^{2} b^{2} + 4 \cdot {(4 a)}^{1} \cdot {(- 3 b)}^{3} + 1 \cdot {(4 a)}^{0} \cdot {(- 3 b)}^{4}$

Simplify.

$256 a^{4} - 768 a^{3} b + 864 a^{2} b^{2} + 4 \cdot (4 a) \cdot {(- 3 b)}^{3} + 1 \cdot {(4 a)}^{0} \cdot {(- 3 b)}^{4}$

Multiply $4$ by $4$ .

$256 a^{4} - 768 a^{3} b + 864 a^{2} b^{2} + 16 a \cdot {(- 3 b)}^{3} + 1 \cdot {(4 a)}^{0} \cdot {(- 3 b)}^{4}$

Apply the product rule to $- 3 b$ .

$256 a^{4} - 768 a^{3} b + 864 a^{2} b^{2} + 16 a \cdot ({(- 3)}^{3} b^{3}) + 1 \cdot {(4 a)}^{0} \cdot {(- 3 b)}^{4}$

Rewrite using the commutative property of multiplication.

$256 a^{4} - 768 a^{3} b + 864 a^{2} b^{2} + 16 \cdot {(- 3)}^{3} a b^{3} + 1 \cdot {(4 a)}^{0} \cdot {(- 3 b)}^{4}$

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

$256 a^{4} - 768 a^{3} b + 864 a^{2} b^{2} + 16 \cdot - 27 a b^{3} + 1 \cdot {(4 a)}^{0} \cdot {(- 3 b)}^{4}$

Multiply $16$ by $- 27$ .

$256 a^{4} - 768 a^{3} b + 864 a^{2} b^{2} - 432 a b^{3} + 1 \cdot {(4 a)}^{0} \cdot {(- 3 b)}^{4}$

Multiply ${(4 a)}^{0}$ by $1$ .

$256 a^{4} - 768 a^{3} b + 864 a^{2} b^{2} - 432 a b^{3} + {(4 a)}^{0} \cdot {(- 3 b)}^{4}$

Apply the product rule to $4 a$ .

$256 a^{4} - 768 a^{3} b + 864 a^{2} b^{2} - 432 a b^{3} + 4^{0} a^{0} \cdot {(- 3 b)}^{4}$

Anything raised to $0$ is $1$ .

$256 a^{4} - 768 a^{3} b + 864 a^{2} b^{2} - 432 a b^{3} + 1 a^{0} \cdot {(- 3 b)}^{4}$

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

$256 a^{4} - 768 a^{3} b + 864 a^{2} b^{2} - 432 a b^{3} + a^{0} \cdot {(- 3 b)}^{4}$

Anything raised to $0$ is $1$ .

$256 a^{4} - 768 a^{3} b + 864 a^{2} b^{2} - 432 a b^{3} + 1 \cdot {(- 3 b)}^{4}$

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

$256 a^{4} - 768 a^{3} b + 864 a^{2} b^{2} - 432 a b^{3} + {(- 3 b)}^{4}$

Apply the product rule to $- 3 b$ .

$256 a^{4} - 768 a^{3} b + 864 a^{2} b^{2} - 432 a b^{3} + {(- 3)}^{4} b^{4}$

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

$256 a^{4} - 768 a^{3} b + 864 a^{2} b^{2} - 432 a b^{3} + 81 b^{4}$

Expand Using the Binomial Theorem (4a-3b)^4

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