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

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

Step 1

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

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

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

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

Step 4

Simplify each term.

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

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

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

$64 \cdot {(- 3 x)}^{0} + 3 \cdot {(4)}^{2} \cdot {(- 3 x)}^{1} + 3 \cdot {(4)}^{1} \cdot {(- 3 x)}^{2} + 1 \cdot {(4)}^{0} \cdot {(- 3 x)}^{3}$

Apply the product rule to $- 3 x$ .

$64 \cdot ({(- 3)}^{0} x^{0}) + 3 \cdot {(4)}^{2} \cdot {(- 3 x)}^{1} + 3 \cdot {(4)}^{1} \cdot {(- 3 x)}^{2} + 1 \cdot {(4)}^{0} \cdot {(- 3 x)}^{3}$

Anything raised to $0$ is $1$ .

$64 \cdot (1 x^{0}) + 3 \cdot {(4)}^{2} \cdot {(- 3 x)}^{1} + 3 \cdot {(4)}^{1} \cdot {(- 3 x)}^{2} + 1 \cdot {(4)}^{0} \cdot {(- 3 x)}^{3}$

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

$64 \cdot x^{0} + 3 \cdot {(4)}^{2} \cdot {(- 3 x)}^{1} + 3 \cdot {(4)}^{1} \cdot {(- 3 x)}^{2} + 1 \cdot {(4)}^{0} \cdot {(- 3 x)}^{3}$

Anything raised to $0$ is $1$ .

$64 \cdot 1 + 3 \cdot {(4)}^{2} \cdot {(- 3 x)}^{1} + 3 \cdot {(4)}^{1} \cdot {(- 3 x)}^{2} + 1 \cdot {(4)}^{0} \cdot {(- 3 x)}^{3}$

Multiply $64$ by $1$ .

$64 + 3 \cdot {(4)}^{2} \cdot {(- 3 x)}^{1} + 3 \cdot {(4)}^{1} \cdot {(- 3 x)}^{2} + 1 \cdot {(4)}^{0} \cdot {(- 3 x)}^{3}$

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

$64 + 3 \cdot 16 \cdot {(- 3 x)}^{1} + 3 \cdot {(4)}^{1} \cdot {(- 3 x)}^{2} + 1 \cdot {(4)}^{0} \cdot {(- 3 x)}^{3}$

Multiply $3$ by $16$ .

$64 + 48 \cdot {(- 3 x)}^{1} + 3 \cdot {(4)}^{1} \cdot {(- 3 x)}^{2} + 1 \cdot {(4)}^{0} \cdot {(- 3 x)}^{3}$

Simplify.

$64 + 48 \cdot (- 3 x) + 3 \cdot {(4)}^{1} \cdot {(- 3 x)}^{2} + 1 \cdot {(4)}^{0} \cdot {(- 3 x)}^{3}$

Multiply $- 3$ by $48$ .

$64 - 144 x + 3 \cdot {(4)}^{1} \cdot {(- 3 x)}^{2} + 1 \cdot {(4)}^{0} \cdot {(- 3 x)}^{3}$

Evaluate the exponent.

$64 - 144 x + 3 \cdot 4 \cdot {(- 3 x)}^{2} + 1 \cdot {(4)}^{0} \cdot {(- 3 x)}^{3}$

Multiply $3$ by $4$ .

$64 - 144 x + 12 \cdot {(- 3 x)}^{2} + 1 \cdot {(4)}^{0} \cdot {(- 3 x)}^{3}$

Apply the product rule to $- 3 x$ .

$64 - 144 x + 12 \cdot ({(- 3)}^{2} x^{2}) + 1 \cdot {(4)}^{0} \cdot {(- 3 x)}^{3}$

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

$64 - 144 x + 12 \cdot (9 x^{2}) + 1 \cdot {(4)}^{0} \cdot {(- 3 x)}^{3}$

Multiply $9$ by $12$ .

$64 - 144 x + 108 x^{2} + 1 \cdot {(4)}^{0} \cdot {(- 3 x)}^{3}$

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

$64 - 144 x + 108 x^{2} + {(4)}^{0} \cdot {(- 3 x)}^{3}$

Anything raised to $0$ is $1$ .

$64 - 144 x + 108 x^{2} + 1 \cdot {(- 3 x)}^{3}$

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

$64 - 144 x + 108 x^{2} + {(- 3 x)}^{3}$

Apply the product rule to $- 3 x$ .

$64 - 144 x + 108 x^{2} + {(- 3)}^{3} x^{3}$

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

$64 - 144 x + 108 x^{2} - 27 x^{3}$

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

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