Find g(f(x)) f(x) = cube root of x-1 , g(x)=x^3+1

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

Set up the composite result function.

$g (f (x))$

Evaluate $g (\sqrt[3]{x - 1})$ by substituting in the value of $f$ into $g$ .

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

Rewrite ${\sqrt[3]{x - 1}}^{3}$ as $x - 1$ .

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

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

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

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

$g (\sqrt[3]{x - 1}) = (x - 1) + 1$

Simplify.

$g (\sqrt[3]{x - 1}) = x - 1 + 1$

Combine the opposite terms in $x - 1 + 1$ .

Add $- 1$ and $1$ .

$g (\sqrt[3]{x - 1}) = x + 0$

Add $x$ and $0$ .

$g (\sqrt[3]{x - 1}) = x$