Solve x^3-3=213

$${ x }^{ 3 } -3=213$$

Solve for x (complex solution)

$$x=-3\sqrt{3}i-3\approx -3-5.196152423i$$

$$x=6$$

$$x=-3+3\sqrt{3}i\approx -3+5.196152423i$$

Solution Steps

Subtract 213 from both sides.

$$x^{3}-3-213=0$$

Subtract 213 from -3 to get -216.

$$x^{3}-216=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -216 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±216,±108,±72,±54,±36,±27,±24,±18,±12,±9,±8,±6,±4,±3,±2,±1$$

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.

$$x=6$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-216 by x-6 to get x^{2}+6x+36. Solve the equation where the result equals to 0.

$$x^{2}+6x+36=0$$

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, 6 for b, and 36 for c in the quadratic formula.

$$x=\frac{-6±\sqrt{6^{2}-4\times 1\times 36}}{2}$$

Do the calculations.

$$x=\frac{-6±\sqrt{-108}}{2}$$

Solve the equation x^{2}+6x+36=0 when ± is plus and when ± is minus.

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

List all found solutions.

$$x=6 x=-3i\sqrt{3}-3 x=-3+3i\sqrt{3}$$

Solve for x

$$x=6$$

Solution Steps

Subtract 213 from both sides.

$$x^{3}-3-213=0$$

Subtract 213 from -3 to get -216.

$$x^{3}-216=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -216 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±216,±108,±72,±54,±36,±27,±24,±18,±12,±9,±8,±6,±4,±3,±2,±1$$

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.

$$x=6$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-216 by x-6 to get x^{2}+6x+36. Solve the equation where the result equals to 0.

$$x^{2}+6x+36=0$$

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, 6 for b, and 36 for c in the quadratic formula.

$$x=\frac{-6±\sqrt{6^{2}-4\times 1\times 36}}{2}$$

Do the calculations.

$$x=\frac{-6±\sqrt{-108}}{2}$$

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.

$$x=6$$