Solve x^3=8000

$$x ^ { 3 } = 8000$$

Solve for x (complex solution)

$$x=-10\sqrt{3}i-10\approx -10-17.320508076i$$

$$x=20$$

$$x=-10+10\sqrt{3}i\approx -10+17.320508076i$$

Solution Steps

Subtract 8000 from both sides.

$$x^{3}-8000=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -8000 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±8000,±4000,±2000,±1600,±1000,±800,±500,±400,±320,±250,±200,±160,±125,±100,±80,±64,±50,±40,±32,±25,±20,±16,±10,±8,±5,±4,±2,±1$$

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.

$$x=20$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-8000 by x-20 to get x^{2}+20x+400. Solve the equation where the result equals to 0.

$$x^{2}+20x+400=0$$

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, 20 for b, and 400 for c in the quadratic formula.

$$x=\frac{-20±\sqrt{20^{2}-4\times 1\times 400}}{2}$$

Do the calculations.

$$x=\frac{-20±\sqrt{-1200}}{2}$$

Solve the equation x^{2}+20x+400=0 when ± is plus and when ± is minus.

$$x=-10i\sqrt{3}-10 x=-10+10i\sqrt{3}$$

List all found solutions.

$$x=20 x=-10i\sqrt{3}-10 x=-10+10i\sqrt{3}$$

Solve for x

$$x=20$$

Solution Steps

Subtract 8000 from both sides.

$$x^{3}-8000=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -8000 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±8000,±4000,±2000,±1600,±1000,±800,±500,±400,±320,±250,±200,±160,±125,±100,±80,±64,±50,±40,±32,±25,±20,±16,±10,±8,±5,±4,±2,±1$$

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.

$$x=20$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-8000 by x-20 to get x^{2}+20x+400. Solve the equation where the result equals to 0.

$$x^{2}+20x+400=0$$

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, 20 for b, and 400 for c in the quadratic formula.

$$x=\frac{-20±\sqrt{20^{2}-4\times 1\times 400}}{2}$$

Do the calculations.

$$x=\frac{-20±\sqrt{-1200}}{2}$$

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.

$$x=20$$