Solve 2(x-1)^3=54

$$2 ( x - 1 ) ^ { 3 } = 54$$

Solve for x (complex solution)

$$x=4$$

$$x=\frac{-1+3\sqrt{3}i}{2}\approx -0.5+2.598076211i$$

$$x=\frac{-3\sqrt{3}i-1}{2}\approx -0.5-2.598076211i$$

Solution Steps

Divide both sides by 2.

$$\left(x-1\right)^{3}=\frac{54}{2}$$

Divide 54 by 2 to get 27.

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

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

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

Subtract 27 from both sides.

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

Subtract 27 from -1 to get -28.

$$x^{3}-3x^{2}+3x-28=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -28 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±28,±14,±7,±4,±2,±1$$

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.

$$x=4$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-3x^{2}+3x-28 by x-4 to get x^{2}+x+7. Solve the equation where the result equals to 0.

$$x^{2}+x+7=0$$

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 7 for c in the quadratic formula.

$$x=\frac{-1±\sqrt{1^{2}-4\times 1\times 7}}{2}$$

Do the calculations.

$$x=\frac{-1±\sqrt{-27}}{2}$$

Solve the equation x^{2}+x+7=0 when ± is plus and when ± is minus.

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

List all found solutions.

$$x=4 x=\frac{-3i\sqrt{3}-1}{2} x=\frac{-1+3i\sqrt{3}}{2}$$

Solve for x

$$x=4$$

Solution Steps

Divide both sides by 2.

$$\left(x-1\right)^{3}=\frac{54}{2}$$

Divide 54 by 2 to get 27.

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

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

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

Subtract 27 from both sides.

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

Subtract 27 from -1 to get -28.

$$x^{3}-3x^{2}+3x-28=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -28 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±28,±14,±7,±4,±2,±1$$

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.

$$x=4$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-3x^{2}+3x-28 by x-4 to get x^{2}+x+7. Solve the equation where the result equals to 0.

$$x^{2}+x+7=0$$

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 7 for c in the quadratic formula.

$$x=\frac{-1±\sqrt{1^{2}-4\times 1\times 7}}{2}$$

Do the calculations.

$$x=\frac{-1±\sqrt{-27}}{2}$$

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.

$$x=4$$