Solve 8x^3+27=0

$$8 x ^ { 3 } + 27 = 0$$

Solve for x (complex solution)

$$x=\frac{-3\sqrt{3}i+3}{4}\approx 0.75-1.299038106i$$

$$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

$$x=\frac{3+3\sqrt{3}i}{4}\approx 0.75+1.299038106i$$

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 27 and q divides the leading coefficient 8. List all candidates \frac{p}{q}.

$$±\frac{27}{8},±\frac{27}{4},±\frac{27}{2},±27,±\frac{9}{8},±\frac{9}{4},±\frac{9}{2},±9,±\frac{3}{8},±\frac{3}{4},±\frac{3}{2},±3,±\frac{1}{8},±\frac{1}{4},±\frac{1}{2},±1$$

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.

$$x=-\frac{3}{2}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 8x^{3}+27 by 2\left(x+\frac{3}{2}\right)=2x+3 to get 4x^{2}-6x+9. Solve the equation where the result equals to 0.

$$4x^{2}-6x+9=0$$

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 4 for a, -6 for b, and 9 for c in the quadratic formula.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 4\times 9}}{2\times 4}$$

Do the calculations.

$$x=\frac{6±\sqrt{-108}}{8}$$

Solve the equation 4x^{2}-6x+9=0 when ± is plus and when ± is minus.

$$x=\frac{-3i\sqrt{3}+3}{4} x=\frac{3+3i\sqrt{3}}{4}$$

List all found solutions.

$$x=-\frac{3}{2} x=\frac{-3i\sqrt{3}+3}{4} x=\frac{3+3i\sqrt{3}}{4}$$

Solve for x

$$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

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 27 and q divides the leading coefficient 8. List all candidates \frac{p}{q}.

$$±\frac{27}{8},±\frac{27}{4},±\frac{27}{2},±27,±\frac{9}{8},±\frac{9}{4},±\frac{9}{2},±9,±\frac{3}{8},±\frac{3}{4},±\frac{3}{2},±3,±\frac{1}{8},±\frac{1}{4},±\frac{1}{2},±1$$

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.

$$x=-\frac{3}{2}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 8x^{3}+27 by 2\left(x+\frac{3}{2}\right)=2x+3 to get 4x^{2}-6x+9. Solve the equation where the result equals to 0.

$$4x^{2}-6x+9=0$$

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 4 for a, -6 for b, and 9 for c in the quadratic formula.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 4\times 9}}{2\times 4}$$

Do the calculations.

$$x=\frac{6±\sqrt{-108}}{8}$$

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.

$$x=-\frac{3}{2}$$