Solve for r cube root of 7r+5=-3

$\sqrt[3]{7 r + 5} = - 3$

Step 1

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

${\sqrt[3]{7 r + 5}}^{3} = {(- 3)}^{3}$

Step 2

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{7 r + 5}$ as ${(7 r + 5)}^{\frac{1}{3}}$ .

${({(7 r + 5)}^{\frac{1}{3}})}^{3} = {(- 3)}^{3}$

Simplify the left side.

Simplify ${({(7 r + 5)}^{\frac{1}{3}})}^{3}$ .

Multiply the exponents in ${({(7 r + 5)}^{\frac{1}{3}})}^{3}$ .

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

${(7 r + 5)}^{\frac{1}{3} \cdot 3} = {(- 3)}^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

${(7 r + 5)}^{\frac{1}{3} \cdot 3} = {(- 3)}^{3}$

Rewrite the expression.

${(7 r + 5)}^{1} = {(- 3)}^{3}$

Simplify.

$7 r + 5 = {(- 3)}^{3}$

Simplify the right side.

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

$7 r + 5 = - 27$

Step 3

Solve for $r$ .

Move all terms not containing $r$ to the right side of the equation.

Subtract $5$ from both sides of the equation.

$7 r = - 27 - 5$

Subtract $5$ from $- 27$ .

$7 r = - 32$

Divide each term in $7 r = - 32$ by $7$ and simplify.

Divide each term in $7 r = - 32$ by $7$ .

$\frac{7 r}{7} = \frac{- 32}{7}$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{7 r}{7} = \frac{- 32}{7}$

Divide $r$ by $1$ .

$r = \frac{- 32}{7}$

Simplify the right side.

Move the negative in front of the fraction.

$r = - \frac{32}{7}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$r = - \frac{32}{7}$

Decimal Form:

$r = - 4.\overline{571428}$

Mixed Number Form:

$r = - 4 \frac{4}{7}$

Solve for r cube root of 7r+5=-3

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