Solve for h 2 cube root of h+5=4 cube root of 2h-15

$2 \sqrt[3]{h + 5} = 4 \sqrt[3]{2 h - 15}$

Step 1

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

${(2 \sqrt[3]{h + 5})}^{3} = {(4 \sqrt[3]{2 h - 15})}^{3}$

Step 2

Simplify each side of the equation.

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

${(2 {(h + 5)}^{\frac{1}{3}})}^{3} = {(4 \sqrt[3]{2 h - 15})}^{3}$

Simplify the left side.

Simplify ${(2 {(h + 5)}^{\frac{1}{3}})}^{3}$ .

Apply the product rule to $2 {(h + 5)}^{\frac{1}{3}}$ .

$2^{3} {({(h + 5)}^{\frac{1}{3}})}^{3} = {(4 \sqrt[3]{2 h - 15})}^{3}$

Raise $2$ to the power of $3$ .

$8 {({(h + 5)}^{\frac{1}{3}})}^{3} = {(4 \sqrt[3]{2 h - 15})}^{3}$

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

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

$8 {(h + 5)}^{\frac{1}{3} \cdot 3} = {(4 \sqrt[3]{2 h - 15})}^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

$8 {(h + 5)}^{\frac{1}{3} \cdot 3} = {(4 \sqrt[3]{2 h - 15})}^{3}$

Rewrite the expression.

$8 {(h + 5)}^{1} = {(4 \sqrt[3]{2 h - 15})}^{3}$

Simplify.

$8 (h + 5) = {(4 \sqrt[3]{2 h - 15})}^{3}$

Apply the distributive property.

$8 h + 8 \cdot 5 = {(4 \sqrt[3]{2 h - 15})}^{3}$

Multiply $8$ by $5$ .

$8 h + 40 = {(4 \sqrt[3]{2 h - 15})}^{3}$

Simplify the right side.

Simplify ${(4 \sqrt[3]{2 h - 15})}^{3}$ .

Simplify the expression.

Apply the product rule to $4 \sqrt[3]{2 h - 15}$ .

$8 h + 40 = 4^{3} {\sqrt[3]{2 h - 15}}^{3}$

Raise $4$ to the power of $3$ .

$8 h + 40 = 64 {\sqrt[3]{2 h - 15}}^{3}$

Rewrite ${\sqrt[3]{2 h - 15}}^{3}$ as $2 h - 15$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{2 h - 15}$ as ${(2 h - 15)}^{\frac{1}{3}}$ .

$8 h + 40 = 64 {({(2 h - 15)}^{\frac{1}{3}})}^{3}$

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

$8 h + 40 = 64 {(2 h - 15)}^{\frac{1}{3} \cdot 3}$

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

$8 h + 40 = 64 {(2 h - 15)}^{\frac{3}{3}}$

Cancel the common factor of $3$ .

Cancel the common factor.

$8 h + 40 = 64 {(2 h - 15)}^{\frac{3}{3}}$

Rewrite the expression.

$8 h + 40 = 64 {(2 h - 15)}^{1}$

Simplify.

$8 h + 40 = 64 (2 h - 15)$

Apply the distributive property.

$8 h + 40 = 64 (2 h) + 64 \cdot - 15$

Multiply.

Multiply $2$ by $64$ .

$8 h + 40 = 128 h + 64 \cdot - 15$

Multiply $64$ by $- 15$ .

$8 h + 40 = 128 h - 960$

Step 3

Solve for $h$ .

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

Subtract $128 h$ from both sides of the equation.

$8 h + 40 - 128 h = - 960$

Subtract $128 h$ from $8 h$ .

$- 120 h + 40 = - 960$

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

Subtract $40$ from both sides of the equation.

$- 120 h = - 960 - 40$

Subtract $40$ from $- 960$ .

$- 120 h = - 1000$

Divide each term in $- 120 h = - 1000$ by $- 120$ and simplify.

Divide each term in $- 120 h = - 1000$ by $- 120$ .

$\frac{- 120 h}{- 120} = \frac{- 1000}{- 120}$

Simplify the left side.

Cancel the common factor of $- 120$ .

Cancel the common factor.

$\frac{- 120 h}{- 120} = \frac{- 1000}{- 120}$

Divide $h$ by $1$ .

$h = \frac{- 1000}{- 120}$

Simplify the right side.

Cancel the common factor of $- 1000$ and $- 120$ .

Factor $- 40$ out of $- 1000$ .

$h = \frac{- 40 (25)}{- 120}$

Cancel the common factors.

Factor $- 40$ out of $- 120$ .

$h = \frac{- 40 \cdot 25}{- 40 \cdot 3}$

Cancel the common factor.

$h = \frac{- 40 \cdot 25}{- 40 \cdot 3}$

Rewrite the expression.

$h = \frac{25}{3}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$h = \frac{25}{3}$

Decimal Form:

$h = 8.\overline{3}$

Mixed Number Form:

$h = 8 \frac{1}{3}$

Solve for h 2 cube root of h+5=4 cube root of 2h-15

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