Solve 36pi=4/3pir_s^3

$$36 \pi = \frac { 4 } { 3 } \pi r _ { s } ^ { 3 }$$

Solve for r_s

$$r_{s}=3$$

Solution Steps

Cancel out \pi on both sides.

$$36=\frac{4}{3}r_{s}^{3}$$

Multiply both sides by \frac{3}{4}, the reciprocal of \frac{4}{3}.

$$36\times \frac{3}{4}=r_{s}^{3}$$

Multiply 36 and \frac{3}{4} to get 27.

$$27=r_{s}^{3}$$

Swap sides so that all variable terms are on the left hand side.

$$r_{s}^{3}=27$$

Subtract 27 from both sides.

$$r_{s}^{3}-27=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -27 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±27,±9,±3,±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$r_{s}=3$$

By Factor theorem, r_{s}-k is a factor of the polynomial for each root k. Divide r_{s}^{3}-27 by r_{s}-3 to get r_{s}^{2}+3r_{s}+9. Solve the equation where the result equals to 0.

$$r_{s}^{2}+3r_{s}+9=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, 3 for b, and 9 for c in the quadratic formula.

$$r_{s}=\frac{-3±\sqrt{3^{2}-4\times 1\times 9}}{2}$$

Do the calculations.

$$r_{s}=\frac{-3±\sqrt{-27}}{2}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$r_{s}\in \emptyset$$

List all found solutions.

$$r_{s}=3$$