Solve for k cube root of k+4 = cube root of 2k+3

$\sqrt[3]{k + 4} = \sqrt[3]{2 k + 3}$

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

${\sqrt[3]{k + 4}}^{3} = {\sqrt[3]{2 k + 3}}^{3}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{k + 4}$ as ${(k + 4)}^{\frac{1}{3}}$ .

${({(k + 4)}^{\frac{1}{3}})}^{3} = {\sqrt[3]{2 k + 3}}^{3}$

Simplify the left side.

Simplify ${({(k + 4)}^{\frac{1}{3}})}^{3}$ .

Multiply the exponents in ${({(k + 4)}^{\frac{1}{3}})}^{3}$ .

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

${(k + 4)}^{\frac{1}{3} \cdot 3} = {\sqrt[3]{2 k + 3}}^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

${(k + 4)}^{\frac{1}{3} \cdot 3} = {\sqrt[3]{2 k + 3}}^{3}$

Rewrite the expression.

${(k + 4)}^{1} = {\sqrt[3]{2 k + 3}}^{3}$

Simplify.

$k + 4 = {\sqrt[3]{2 k + 3}}^{3}$

Simplify the right side.

Rewrite ${\sqrt[3]{2 k + 3}}^{3}$ as $2 k + 3$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{2 k + 3}$ as ${(2 k + 3)}^{\frac{1}{3}}$ .

$k + 4 = {({(2 k + 3)}^{\frac{1}{3}})}^{3}$

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

$k + 4 = {(2 k + 3)}^{\frac{1}{3} \cdot 3}$

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

$k + 4 = {(2 k + 3)}^{\frac{3}{3}}$

Cancel the common factor of $3$ .

Cancel the common factor.

$k + 4 = {(2 k + 3)}^{\frac{3}{3}}$

Rewrite the expression.

$k + 4 = {(2 k + 3)}^{1}$

Simplify.

$k + 4 = 2 k + 3$

Solve for $k$ .

Move all terms containing $k$ to the left side of the equation.

Subtract $2 k$ from both sides of the equation.

$k + 4 - 2 k = 3$

Subtract $2 k$ from $k$ .

$- k + 4 = 3$

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

Subtract $4$ from both sides of the equation.

$- k = 3 - 4$

Subtract $4$ from $3$ .

$- k = - 1$

Divide each term in $- k = - 1$ by $- 1$ and simplify.

Divide each term in $- k = - 1$ by $- 1$ .

$\frac{- k}{- 1} = \frac{- 1}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{k}{1} = \frac{- 1}{- 1}$

Divide $k$ by $1$ .

$k = \frac{- 1}{- 1}$

Simplify the right side.

Divide $- 1$ by $- 1$ .

$k = 1$

Solve for k cube root of k+4 = cube root of 2k+3

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