Solve for j cube root of 7j-2 = cube root of j+4

$\sqrt[3]{7 j - 2} = \sqrt[3]{j + 4}$

Step 1

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

${\sqrt[3]{7 j - 2}}^{3} = {\sqrt[3]{j + 4}}^{3}$

Step 2

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{7 j - 2}$ as ${(7 j - 2)}^{\frac{1}{3}}$ .

${({(7 j - 2)}^{\frac{1}{3}})}^{3} = {\sqrt[3]{j + 4}}^{3}$

Simplify the left side.

Simplify ${({(7 j - 2)}^{\frac{1}{3}})}^{3}$ .

Multiply the exponents in ${({(7 j - 2)}^{\frac{1}{3}})}^{3}$ .

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

${(7 j - 2)}^{\frac{1}{3} \cdot 3} = {\sqrt[3]{j + 4}}^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

${(7 j - 2)}^{\frac{1}{3} \cdot 3} = {\sqrt[3]{j + 4}}^{3}$

Rewrite the expression.

${(7 j - 2)}^{1} = {\sqrt[3]{j + 4}}^{3}$

Simplify.

$7 j - 2 = {\sqrt[3]{j + 4}}^{3}$

Simplify the right side.

Rewrite ${\sqrt[3]{j + 4}}^{3}$ as $j + 4$ .

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

$7 j - 2 = {({(j + 4)}^{\frac{1}{3}})}^{3}$

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

$7 j - 2 = {(j + 4)}^{\frac{1}{3} \cdot 3}$

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

$7 j - 2 = {(j + 4)}^{\frac{3}{3}}$

Cancel the common factor of $3$ .

Cancel the common factor.

$7 j - 2 = {(j + 4)}^{\frac{3}{3}}$

Rewrite the expression.

$7 j - 2 = {(j + 4)}^{1}$

Simplify.

$7 j - 2 = j + 4$

Step 3

Solve for $j$ .

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

Subtract $j$ from both sides of the equation.

$7 j - 2 - j = 4$

Subtract $j$ from $7 j$ .

$6 j - 2 = 4$

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

Add $2$ to both sides of the equation.

$6 j = 4 + 2$

Add $4$ and $2$ .

$6 j = 6$

Divide each term in $6 j = 6$ by $6$ and simplify.

Divide each term in $6 j = 6$ by $6$ .

$\frac{6 j}{6} = \frac{6}{6}$

Simplify the left side.

Cancel the common factor of $6$ .

Cancel the common factor.

$\frac{6 j}{6} = \frac{6}{6}$

Divide $j$ by $1$ .

$j = \frac{6}{6}$

Simplify the right side.

Divide $6$ by $6$ .

$j = 1$

Solve for j cube root of 7j-2 = cube root of j+4

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