Expand the Trigonometric Expression (h+3k)^3

${(h + 3 k)}^{3}$

Step 1

Use the Binomial Theorem.

$h^{3} + 3 h^{2} (3 k) + 3 h {(3 k)}^{2} + {(3 k)}^{3}$

Step 2

Simplify each term.

Rewrite using the commutative property of multiplication.

$h^{3} + 3 \cdot 3 h^{2} k + 3 h {(3 k)}^{2} + {(3 k)}^{3}$

Multiply $3$ by $3$ .

$h^{3} + 9 h^{2} k + 3 h {(3 k)}^{2} + {(3 k)}^{3}$

Apply the product rule to $3 k$ .

$h^{3} + 9 h^{2} k + 3 h (3^{2} k^{2}) + {(3 k)}^{3}$

Rewrite using the commutative property of multiplication.

$h^{3} + 9 h^{2} k + 3 \cdot 3^{2} h k^{2} + {(3 k)}^{3}$

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

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

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

$h^{3} + 9 h^{2} k + 3^{1} \cdot 3^{2} h k^{2} + {(3 k)}^{3}$

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

$h^{3} + 9 h^{2} k + 3^{1 + 2} h k^{2} + {(3 k)}^{3}$

Add $1$ and $2$ .

$h^{3} + 9 h^{2} k + 3^{3} h k^{2} + {(3 k)}^{3}$

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

$h^{3} + 9 h^{2} k + 27 h k^{2} + {(3 k)}^{3}$

Apply the product rule to $3 k$ .

$h^{3} + 9 h^{2} k + 27 h k^{2} + 3^{3} k^{3}$

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

$h^{3} + 9 h^{2} k + 27 h k^{2} + 27 k^{3}$