Solve for r 36pi=4/3pir^3

$36 π = \frac{4}{3} π r^{3}$

Rewrite the equation as $\frac{4}{3} \cdot (π r^{3}) = 36 π$ .

$\frac{4}{3} \cdot (π r^{3}) = 36 π$

Multiply both sides of the equation by $\frac{1}{\frac{4}{3} π}$ .

$\frac{1}{\frac{4}{3} π} (\frac{4}{3} \cdot (π r^{3})) = \frac{1}{\frac{4}{3} π} (36 π)$

Simplify both sides of the equation.

Simplify the left side.

Simplify $\frac{1}{\frac{4}{3} π} (\frac{4}{3} \cdot (π r^{3}))$ .

Combine $\frac{4}{3}$ and $π$ .

$\frac{1}{\frac{4 π}{3}} (\frac{4}{3} \cdot (π r^{3})) = \frac{1}{\frac{4}{3} π} (36 π)$

Multiply the numerator by the reciprocal of the denominator.

$1 \frac{3}{4 π} (\frac{4}{3} \cdot (π r^{3})) = \frac{1}{\frac{4}{3} π} (36 π)$

Multiply $\frac{3}{4 π}$ by $1$ .

$\frac{3}{4 π} (\frac{4}{3} \cdot (π r^{3})) = \frac{1}{\frac{4}{3} π} (36 π)$

Cancel the common factor of $π$ .

Factor $π$ out of $4 π$ .

$\frac{3}{π \cdot 4} (\frac{4}{3} \cdot (π r^{3})) = \frac{1}{\frac{4}{3} π} (36 π)$

Factor $π$ out of $\frac{4}{3} \cdot (π r^{3})$ .

$\frac{3}{π \cdot 4} (π (\frac{4}{3} \cdot (r^{3}))) = \frac{1}{\frac{4}{3} π} (36 π)$

Cancel the common factor.

$\frac{3}{π \cdot 4} (π (\frac{4}{3} \cdot (r^{3}))) = \frac{1}{\frac{4}{3} π} (36 π)$

Rewrite the expression.

$\frac{3}{4} (\frac{4}{3} \cdot (r^{3})) = \frac{1}{\frac{4}{3} π} (36 π)$

Combine $\frac{4}{3}$ and $r^{3}$ .

$\frac{3}{4} \cdot \frac{4 r^{3}}{3} = \frac{1}{\frac{4}{3} π} (36 π)$

Multiply $\frac{3}{4}$ by $\frac{4 r^{3}}{3}$ .

$\frac{3 (4 r^{3})}{4 \cdot 3} = \frac{1}{\frac{4}{3} π} (36 π)$

Multiply.

Multiply $4$ by $3$ .

$\frac{12 r^{3}}{4 \cdot 3} = \frac{1}{\frac{4}{3} π} (36 π)$

Multiply $4$ by $3$ .

$\frac{12 r^{3}}{12} = \frac{1}{\frac{4}{3} π} (36 π)$

Cancel the common factor of $12$ .

Cancel the common factor.

$\frac{12 r^{3}}{12} = \frac{1}{\frac{4}{3} π} (36 π)$

Divide $r^{3}$ by $1$ .

$r^{3} = \frac{1}{\frac{4}{3} π} (36 π)$

Simplify the right side.

Simplify $\frac{1}{\frac{4}{3} π} (36 π)$ .

Cancel the common factor of $π$ .

Factor $π$ out of $\frac{4}{3} π$ .

$r^{3} = \frac{1}{π \frac{4}{3}} (36 π)$

Factor $π$ out of $36 π$ .

$r^{3} = \frac{1}{π \frac{4}{3}} (π \cdot 36)$

Cancel the common factor.

$r^{3} = \frac{1}{π \frac{4}{3}} (π \cdot 36)$

Rewrite the expression.

$r^{3} = \frac{1}{\frac{4}{3}} \cdot 36$

Combine $\frac{1}{\frac{4}{3}}$ and $36$ .

$r^{3} = \frac{36}{\frac{4}{3}}$

Multiply the numerator by the reciprocal of the denominator.

$r^{3} = 36 (\frac{3}{4})$

Cancel the common factor of $4$ .

Factor $4$ out of $36$ .

$r^{3} = 4 (9) \frac{3}{4}$

Cancel the common factor.

$r^{3} = 4 \cdot 9 \frac{3}{4}$

Rewrite the expression.

$r^{3} = 9 \cdot 3$

Multiply $9$ by $3$ .

$r^{3} = 27$

Subtract $27$ from both sides of the equation.

$r^{3} - 27 = 0$

Factor the left side of the equation.

Rewrite $27$ as $3^{3}$ .

$r^{3} - 3^{3} = 0$

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = r$ and $b = 3$ .

$(r - 3) (r^{2} + r \cdot 3 + 3^{2}) = 0$

Simplify.

Move $3$ to the left of $r$ .

$(r - 3) (r^{2} + 3 r + 3^{2}) = 0$

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

$(r - 3) (r^{2} + 3 r + 9) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$r - 3 = 0$

$r^{2} + 3 r + 9 = 0$

Set $r - 3$ equal to $0$ and solve for $r$ .

Set $r - 3$ equal to $0$ .

$r - 3 = 0$

Add $3$ to both sides of the equation.

$r = 3$

Set $r^{2} + 3 r + 9$ equal to $0$ and solve for $r$ .

Set $r^{2} + 3 r + 9$ equal to $0$ .

$r^{2} + 3 r + 9 = 0$

Solve $r^{2} + 3 r + 9 = 0$ for $r$ .

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = 3$ , and $c = 9$ into the quadratic formula and solve for $r$ .

$\frac{- 3 \pm \sqrt{3^{2} - 4 \cdot (1 \cdot 9)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

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

$r = \frac{- 3 \pm \sqrt{9 - 4 \cdot 1 \cdot 9}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$r = \frac{- 3 \pm \sqrt{9 - 4 \cdot 9}}{2 \cdot 1}$

Multiply $- 4$ by $9$ .

$r = \frac{- 3 \pm \sqrt{9 - 36}}{2 \cdot 1}$

Subtract $36$ from $9$ .

$r = \frac{- 3 \pm \sqrt{- 27}}{2 \cdot 1}$

Rewrite $- 27$ as $- 1 (27)$ .

$r = \frac{- 3 \pm \sqrt{- 1 \cdot 27}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (27)}$ as $\sqrt{- 1} \cdot \sqrt{27}$ .

$r = \frac{- 3 \pm \sqrt{- 1} \cdot \sqrt{27}}{2 \cdot 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$r = \frac{- 3 \pm i \cdot \sqrt{27}}{2 \cdot 1}$

Rewrite $27$ as $3^{2} \cdot 3$ .

Factor $9$ out of $27$ .

$r = \frac{- 3 \pm i \cdot \sqrt{9 (3)}}{2 \cdot 1}$

Rewrite $9$ as $3^{2}$ .

$r = \frac{- 3 \pm i \cdot \sqrt{3^{2} \cdot 3}}{2 \cdot 1}$

Pull terms out from under the radical.

$r = \frac{- 3 \pm i \cdot (3 \sqrt{3})}{2 \cdot 1}$

Move $3$ to the left of $i$ .

$r = \frac{- 3 \pm 3 i \sqrt{3}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$r = \frac{- 3 \pm 3 i \sqrt{3}}{2}$

The final answer is the combination of both solutions.

$r = - \frac{3 - 3 i \sqrt{3}}{2}, - \frac{3 + 3 i \sqrt{3}}{2}$

The final solution is all the values that make $(r - 3) (r^{2} + 3 r + 9) = 0$ true.

$r = 3, - \frac{3 - 3 i \sqrt{3}}{2}, - \frac{3 + 3 i \sqrt{3}}{2}$

Solve for r 36pi=4/3pir^3

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