Solve 3-3x=2-2x^3

$$3-3x=2-2 { x }^{ 3 }$$

Solve for x

$$x=\frac{-\sqrt{3}-1}{2}\approx -1.366025404$$

$$x=1$$

$$x=\frac{\sqrt{3}-1}{2}\approx 0.366025404$$

Solution Steps

Subtract 2 from both sides.

$$3-3x-2=-2x^{3}$$

Subtract 2 from 3 to get 1.

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

Add 2x^{3} to both sides.

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

Rearrange the equation to put it in standard form. Place the terms in order from highest to lowest power.

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

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 2. List all candidates \frac{p}{q}.

$$±\frac{1}{2},±1$$

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.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{3}-3x+1 by x-1 to get 2x^{2}+2x-1. Solve the equation where the result equals to 0.

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

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 2 for a, 2 for b, and -1 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{-2±2\sqrt{3}}{4}$$

Solve the equation 2x^{2}+2x-1=0 when ± is plus and when ± is minus.

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

List all found solutions.

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