Simplify (3x+4)^3

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

Step 1

Use the Binomial Theorem.

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

Step 2

Simplify each term.

Apply the product rule to $3 x$ .

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

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

$27 x^{3} + 3 {(3 x)}^{2} \cdot 4 + 3 (3 x) \cdot 4^{2} + 4^{3}$

Apply the product rule to $3 x$ .

$27 x^{3} + 3 (3^{2} x^{2}) \cdot 4 + 3 (3 x) \cdot 4^{2} + 4^{3}$

Multiply $3$ by $3^{2}$ by adding the exponents.

Move $3^{2}$ .

$27 x^{3} + 3^{2} \cdot 3 x^{2} \cdot 4 + 3 (3 x) \cdot 4^{2} + 4^{3}$

Multiply $3^{2}$ by $3$ .

Raise $3$ to the power of $1$ .

$27 x^{3} + 3^{2} \cdot 3^{1} x^{2} \cdot 4 + 3 (3 x) \cdot 4^{2} + 4^{3}$

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

$27 x^{3} + 3^{2 + 1} x^{2} \cdot 4 + 3 (3 x) \cdot 4^{2} + 4^{3}$

Add $2$ and $1$ .

$27 x^{3} + 3^{3} x^{2} \cdot 4 + 3 (3 x) \cdot 4^{2} + 4^{3}$

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

$27 x^{3} + 27 x^{2} \cdot 4 + 3 (3 x) \cdot 4^{2} + 4^{3}$

Multiply $4$ by $27$ .

$27 x^{3} + 108 x^{2} + 3 (3 x) \cdot 4^{2} + 4^{3}$

Multiply $3$ by $3$ .

$27 x^{3} + 108 x^{2} + 9 x \cdot 4^{2} + 4^{3}$

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

$27 x^{3} + 108 x^{2} + 9 x \cdot 16 + 4^{3}$

Multiply $16$ by $9$ .

$27 x^{3} + 108 x^{2} + 144 x + 4^{3}$

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

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