Solve left(x-1right)^3=-27

$${ \left(x-1 \right) }^{ 3 } =-27$$

Solve for x (complex solution)

$$x=-2$$

$$x=\frac{5+3\sqrt{3}i}{2}\approx 2.5+2.598076211i$$

$$x=\frac{-3\sqrt{3}i+5}{2}\approx 2.5-2.598076211i$$

Solution Steps

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(x-1\right)^{3}.

$$x^{3}-3x^{2}+3x-1=-27$$

Add 27 to both sides.

$$x^{3}-3x^{2}+3x-1+27=0$$

Add -1 and 27 to get 26.

$$x^{3}-3x^{2}+3x+26=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 26 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±26,±13,±2,±1$$

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.

$$x=-2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-3x^{2}+3x+26 by x+2 to get x^{2}-5x+13. Solve the equation where the result equals to 0.

$$x^{2}-5x+13=0$$

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, -5 for b, and 13 for c in the quadratic formula.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 1\times 13}}{2}$$

Do the calculations.

$$x=\frac{5±\sqrt{-27}}{2}$$

Solve the equation x^{2}-5x+13=0 when ± is plus and when ± is minus.

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

List all found solutions.

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

Solve for x

$$x=-2$$

Solution Steps

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(x-1\right)^{3}.

$$x^{3}-3x^{2}+3x-1=-27$$

Add 27 to both sides.

$$x^{3}-3x^{2}+3x-1+27=0$$

Add -1 and 27 to get 26.

$$x^{3}-3x^{2}+3x+26=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 26 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±26,±13,±2,±1$$

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.

$$x=-2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-3x^{2}+3x+26 by x+2 to get x^{2}-5x+13. Solve the equation where the result equals to 0.

$$x^{2}-5x+13=0$$

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, -5 for b, and 13 for c in the quadratic formula.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 1\times 13}}{2}$$

Do the calculations.

$$x=\frac{5±\sqrt{-27}}{2}$$

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.

$$x=-2$$