Solve (x-1)^3=64

$$( x - 1 ) ^ { 3 } = 64$$

Solve for x (complex solution)

$$x=5$$

$$x=-1+2\sqrt{3}i\approx -1+3.464101615i$$

$$x=-2\sqrt{3}i-1\approx -1-3.464101615i$$

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}.

$$x^{3}-3x^{2}+3x-1=64$$

Subtract 64 from both sides.

$$x^{3}-3x^{2}+3x-1-64=0$$

Subtract 64 from -1 to get -65.

$$x^{3}-3x^{2}+3x-65=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -65 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±65,±13,±5,±1$$

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.

$$x=5$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-3x^{2}+3x-65 by x-5 to get x^{2}+2x+13. Solve the equation where the result equals to 0.

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

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, 2 for b, and 13 for c in the quadratic formula.

$$x=\frac{-2±\sqrt{2^{2}-4\times 1\times 13}}{2}$$

Do the calculations.

$$x=\frac{-2±\sqrt{-48}}{2}$$

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

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

List all found solutions.

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

Solve for x

$$x=5$$

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}.

$$x^{3}-3x^{2}+3x-1=64$$

Subtract 64 from both sides.

$$x^{3}-3x^{2}+3x-1-64=0$$

Subtract 64 from -1 to get -65.

$$x^{3}-3x^{2}+3x-65=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -65 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±65,±13,±5,±1$$

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.

$$x=5$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-3x^{2}+3x-65 by x-5 to get x^{2}+2x+13. Solve the equation where the result equals to 0.

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

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, 2 for b, and 13 for c in the quadratic formula.

$$x=\frac{-2±\sqrt{2^{2}-4\times 1\times 13}}{2}$$

Do the calculations.

$$x=\frac{-2±\sqrt{-48}}{2}$$

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.

$$x=5$$