Solve 4/3pix^3=36pi

$$\frac{ 4 }{ 3 } \pi { x }^{ 3 } =36 \pi$$

Solve for x (complex solution)

$$x=\frac{-3\sqrt{3}i-3}{2}\approx -1.5-2.598076211i$$

$$x=3$$

$$x=\frac{-3+3\sqrt{3}i}{2}\approx -1.5+2.598076211i$$

Solution Steps

Cancel out \pi on both sides.

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

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

$$x^{3}=36\times \frac{3}{4}$$

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

$$x^{3}=27$$

Subtract 27 from both sides.

$$x^{3}-27=0$$

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.

$$x=3$$

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

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

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.

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

Do the calculations.

$$x=\frac{-3±\sqrt{-27}}{2}$$

Solve the equation x^{2}+3x+9=0 when ± is plus and when ± is minus.

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

List all found solutions.

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

Solve for x

$$x=3$$

Solution Steps

Cancel out \pi on both sides.

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

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

$$x^{3}=36\times \frac{3}{4}$$

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

$$x^{3}=27$$

Subtract 27 from both sides.

$$x^{3}-27=0$$

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.

$$x=3$$

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

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

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.

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

Do the calculations.

$$x=\frac{-3±\sqrt{-27}}{2}$$

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

$$x\in \emptyset$$

List all found solutions.

$$x=3$$