Solve (x-3)^3-125=0

$$( x - 3 ) ^ { 3 } - 125 = 0$$

Solve for x (complex solution)

$$x=8$$

$$x=\frac{1+5\sqrt{3}i}{2}\approx 0.5+4.330127019i$$

$$x=\frac{-5\sqrt{3}i+1}{2}\approx 0.5-4.330127019i$$

Solution Steps

Expand the expression.

$$x^{3}-9x^{2}+27x-152=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -152 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±152,±76,±38,±19,±8,±4,±2,±1$$

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.

$$x=8$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-9x^{2}+27x-152 by x-8 to get x^{2}-x+19. Solve the equation where the result equals to 0.

$$x^{2}-x+19=0$$

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, -1 for b, and 19 for c in the quadratic formula.

$$x=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 1\times 19}}{2}$$

Do the calculations.

$$x=\frac{1±\sqrt{-75}}{2}$$

Solve the equation x^{2}-x+19=0 when ± is plus and when ± is minus.

$$x=\frac{-5i\sqrt{3}+1}{2} x=\frac{1+5i\sqrt{3}}{2}$$

List all found solutions.

$$x=8 x=\frac{-5i\sqrt{3}+1}{2} x=\frac{1+5i\sqrt{3}}{2}$$

Solve for x

$$x=8$$

Solution Steps

Expand the expression.

$$x^{3}-9x^{2}+27x-152=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -152 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±152,±76,±38,±19,±8,±4,±2,±1$$

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.

$$x=8$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-9x^{2}+27x-152 by x-8 to get x^{2}-x+19. Solve the equation where the result equals to 0.

$$x^{2}-x+19=0$$

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, -1 for b, and 19 for c in the quadratic formula.

$$x=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 1\times 19}}{2}$$

Do the calculations.

$$x=\frac{1±\sqrt{-75}}{2}$$

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.

$$x=8$$