Solve for b cube root of b+5=4

$\sqrt[3]{b} + 5 = 4$

Step 1

Move all terms not containing $\sqrt[3]{b}$ to the right side of the equation.

Subtract $5$ from both sides of the equation.

$\sqrt[3]{b} = 4 - 5$

Subtract $5$ from $4$ .

$\sqrt[3]{b} = - 1$

Step 2

To remove the radical on the left side of the equation, cube both sides of the equation.

${\sqrt[3]{b}}^{3} = {(- 1)}^{3}$

Step 3

Simplify each side of the equation.

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

${(b^{\frac{1}{3}})}^{3} = {(- 1)}^{3}$

Simplify the left side.

Simplify ${(b^{\frac{1}{3}})}^{3}$ .

Multiply the exponents in ${(b^{\frac{1}{3}})}^{3}$ .

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

$b^{\frac{1}{3} \cdot 3} = {(- 1)}^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

$b^{\frac{1}{3} \cdot 3} = {(- 1)}^{3}$

Rewrite the expression.

$b^{1} = {(- 1)}^{3}$

Simplify.

$b = {(- 1)}^{3}$

Simplify the right side.

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

$b = - 1$