Solve for y cube root of 19y+40=3 cube root of y

$\sqrt[3]{19 y + 40} = 3 \sqrt[3]{y}$

Step 1

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

${\sqrt[3]{19 y + 40}}^{3} = {(3 \sqrt[3]{y})}^{3}$

Step 2

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{19 y + 40}$ as ${(19 y + 40)}^{\frac{1}{3}}$ .

${({(19 y + 40)}^{\frac{1}{3}})}^{3} = {(3 \sqrt[3]{y})}^{3}$

Simplify the left side.

Simplify ${({(19 y + 40)}^{\frac{1}{3}})}^{3}$ .

Multiply the exponents in ${({(19 y + 40)}^{\frac{1}{3}})}^{3}$ .

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

${(19 y + 40)}^{\frac{1}{3} \cdot 3} = {(3 \sqrt[3]{y})}^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

${(19 y + 40)}^{\frac{1}{3} \cdot 3} = {(3 \sqrt[3]{y})}^{3}$

Rewrite the expression.

${(19 y + 40)}^{1} = {(3 \sqrt[3]{y})}^{3}$

Simplify.

$19 y + 40 = {(3 \sqrt[3]{y})}^{3}$

Simplify the right side.

Simplify ${(3 \sqrt[3]{y})}^{3}$ .

Simplify the expression.

Apply the product rule to $3 \sqrt[3]{y}$ .

$19 y + 40 = 3^{3} {\sqrt[3]{y}}^{3}$

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

$19 y + 40 = 27 {\sqrt[3]{y}}^{3}$

Rewrite ${\sqrt[3]{y}}^{3}$ as $y$ .

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

$19 y + 40 = 27 {(y^{\frac{1}{3}})}^{3}$

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

$19 y + 40 = 27 y^{\frac{1}{3} \cdot 3}$

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

$19 y + 40 = 27 y^{\frac{3}{3}}$

Cancel the common factor of $3$ .

Cancel the common factor.

$19 y + 40 = 27 y^{\frac{3}{3}}$

Rewrite the expression.

$19 y + 40 = 27 y^{1}$

Simplify.

$19 y + 40 = 27 y$

Step 3

Solve for $y$ .

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

Subtract $27 y$ from both sides of the equation.

$19 y + 40 - 27 y = 0$

Subtract $27 y$ from $19 y$ .

$- 8 y + 40 = 0$

Subtract $40$ from both sides of the equation.

$- 8 y = - 40$

Divide each term in $- 8 y = - 40$ by $- 8$ and simplify.

Divide each term in $- 8 y = - 40$ by $- 8$ .

$\frac{- 8 y}{- 8} = \frac{- 40}{- 8}$

Simplify the left side.

Cancel the common factor of $- 8$ .

Cancel the common factor.

$\frac{- 8 y}{- 8} = \frac{- 40}{- 8}$

Divide $y$ by $1$ .

$y = \frac{- 40}{- 8}$

Simplify the right side.

Divide $- 40$ by $- 8$ .

$y = 5$

Solve for y cube root of 19y+40=3 cube root of y

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