Algebra Examples

$f (x) = \sqrt[3]{x - 2}$ , $g (x) = x^{3} + 1$

Step 1

Set up the composite result function.

$f (g (x))$

Step 2

Evaluate $f (x^{3} + 1)$ by substituting in the value of $g$ into $f$ .

$f (x^{3} + 1) = \sqrt[3]{(x^{3} + 1) - 2}$

Step 3

Simplify the expression.

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Step 3.1

Subtract $2$ from $1$ .

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

Step 3.2

Rewrite $1$ as $1^{3}$ .

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

Step 4

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = x$ and $b = 1$ .

$f (x^{3} + 1) = \sqrt[3]{(x - 1) (x^{2} + x \cdot 1 + 1^{2})}$

Step 5

Simplify.

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Step 5.1

Multiply $x$ by $1$ .

$f (x^{3} + 1) = \sqrt[3]{(x - 1) (x^{2} + x + 1^{2})}$

Step 5.2

One to any power is one.

$f (x^{3} + 1) = \sqrt[3]{(x - 1) (x^{2} + x + 1)}$