Solve 27x^3+1=0

$$27 x ^ { 3 } + 1 = 0$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{3}i+1}{6}\approx 0.166666667-0.288675135i$$

$$x=-\frac{1}{3}\approx -0.333333333$$

$$x=\frac{1+\sqrt{3}i}{6}\approx 0.166666667+0.288675135i$$

Solution Steps

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

$$±\frac{1}{27},±\frac{1}{9},±\frac{1}{3},±1$$

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.

$$x=-\frac{1}{3}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 27x^{3}+1 by 3\left(x+\frac{1}{3}\right)=3x+1 to get 9x^{2}-3x+1. Solve the equation where the result equals to 0.

$$9x^{2}-3x+1=0$$

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 9 for a, -3 for b, and 1 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{3±\sqrt{-27}}{18}$$

Solve the equation 9x^{2}-3x+1=0 when ± is plus and when ± is minus.

$$x=\frac{-\sqrt{3}i+1}{6} x=\frac{1+\sqrt{3}i}{6}$$

List all found solutions.

$$x=-\frac{1}{3} x=\frac{-\sqrt{3}i+1}{6} x=\frac{1+\sqrt{3}i}{6}$$

Solve for x

$$x=-\frac{1}{3}\approx -0.333333333$$

Solution Steps

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

$$±\frac{1}{27},±\frac{1}{9},±\frac{1}{3},±1$$

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.

$$x=-\frac{1}{3}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 27x^{3}+1 by 3\left(x+\frac{1}{3}\right)=3x+1 to get 9x^{2}-3x+1. Solve the equation where the result equals to 0.

$$9x^{2}-3x+1=0$$

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 9 for a, -3 for b, and 1 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{3±\sqrt{-27}}{18}$$

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.

$$x=-\frac{1}{3}$$