Find h(f(x)) f(x)=(3x-5)^3 h(x)=2 cube root of x+8

$f (x) = {(3 x - 5)}^{3}$ $h (x) = 2 \sqrt[3]{x} + 8$

Set up the composite result function.

$h (f (x))$

Evaluate $h ({(3 x - 5)}^{3})$ by substituting in the value of $f$ into $h$ .

$h ({(3 x - 5)}^{3}) = 2 \sqrt[3]{{(3 x - 5)}^{3}} + 8$

Remove parentheses.

$h ({(3 x - 5)}^{3}) = 2 \sqrt[3]{{(3 x - 5)}^{3}} + 8$

Simplify each term.

Pull terms out from under the radical, assuming real numbers.

$h ({(3 x - 5)}^{3}) = 2 (3 x - 5) + 8$

Apply the distributive property.

$h ({(3 x - 5)}^{3}) = 2 (3 x) + 2 \cdot - 5 + 8$

Multiply $3$ by $2$ .

$h ({(3 x - 5)}^{3}) = 6 x + 2 \cdot - 5 + 8$

Multiply $2$ by $- 5$ .

$h ({(3 x - 5)}^{3}) = 6 x - 10 + 8$

Add $- 10$ and $8$ .

$h ({(3 x - 5)}^{3}) = 6 x - 2$