Solve 0=(2x-1)^3

$$0 = ( 2 x - 1 ) ^ { 3 }$$

Solve for x

$$x=\frac{1}{2}=0.5$$

Solution Steps

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

$$0=8x^{3}-12x^{2}+6x-1$$

Swap sides so that all variable terms are on the left hand side.

$$8x^{3}-12x^{2}+6x-1=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -1 and q divides the leading coefficient 8. List all candidates \frac{p}{q}.

$$±\frac{1}{8},±\frac{1}{4},±\frac{1}{2},±1$$

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.

$$x=\frac{1}{2}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 8x^{3}-12x^{2}+6x-1 by 2\left(x-\frac{1}{2}\right)=2x-1 to get 4x^{2}-4x+1. Solve the equation where the result equals to 0.

$$4x^{2}-4x+1=0$$

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, -4 for b, and 1 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{4±0}{8}$$

Solutions are the same.

$$x=\frac{1}{2}$$