Find f(g(x)) f(x)=x^3-3 g(x) = cube root of x+3

$f (x) = x^{3} - 3$ $g (x) = \sqrt[3]{x + 3}$

Step 1

Set up the composite result function.

$f (g (x))$

Step 2

Evaluate $f (\sqrt[3]{x + 3})$ by substituting in the value of $g$ into $f$ .

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

Step 3

Rewrite ${\sqrt[3]{x + 3}}^{3}$ as $x + 3$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{x + 3}$ as ${(x + 3)}^{\frac{1}{3}}$ .

$f (\sqrt[3]{x + 3}) = {({(x + 3)}^{\frac{1}{3}})}^{3} - 3$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f (\sqrt[3]{x + 3}) = {(x + 3)}^{\frac{1}{3} \cdot 3} - 3$

Combine $\frac{1}{3}$ and $3$ .

$f (\sqrt[3]{x + 3}) = {(x + 3)}^{\frac{3}{3}} - 3$

Cancel the common factor of $3$ .

Cancel the common factor.

$f (\sqrt[3]{x + 3}) = {(x + 3)}^{\frac{3}{3}} - 3$

Rewrite the expression.

$f (\sqrt[3]{x + 3}) = (x + 3) - 3$

Simplify.

$f (\sqrt[3]{x + 3}) = x + 3 - 3$

Step 4

Combine the opposite terms in $x + 3 - 3$ .

Subtract $3$ from $3$ .

$f (\sqrt[3]{x + 3}) = x + 0$

Add $x$ and $0$ .

$f (\sqrt[3]{x + 3}) = x$