Solve left(2-xright)^3-left(2-xright)^2=3-4x^3/4

$${ \left(2-x \right) }^{ 3 } - { \left(2-x \right) }^{ 2 } = \frac{ 3-4 { x }^{ 3 } }{ 4 }$$

Solve for x (complex solution)

$$x=\frac{4}{5}+\frac{1}{10}i=0.8+0.1i$$

$$x=\frac{4}{5}-\frac{1}{10}i=0.8-0.1i$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 4.

$$4\left(2-x\right)^{3}-4\left(2-x\right)^{2}=3-4x^{3}$$

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

$$4\left(8-12x+6x^{2}-x^{3}\right)-4\left(2-x\right)^{2}=3-4x^{3}$$

Use the distributive property to multiply 4 by 8-12x+6x^{2}-x^{3}.

$$32-48x+24x^{2}-4x^{3}-4\left(2-x\right)^{2}=3-4x^{3}$$

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

$$32-48x+24x^{2}-4x^{3}-4\left(4-4x+x^{2}\right)=3-4x^{3}$$

Use the distributive property to multiply -4 by 4-4x+x^{2}.

$$32-48x+24x^{2}-4x^{3}-16+16x-4x^{2}=3-4x^{3}$$

Subtract 16 from 32 to get 16.

$$16-48x+24x^{2}-4x^{3}+16x-4x^{2}=3-4x^{3}$$

Combine -48x and 16x to get -32x.

$$16-32x+24x^{2}-4x^{3}-4x^{2}=3-4x^{3}$$

Combine 24x^{2} and -4x^{2} to get 20x^{2}.

$$16-32x+20x^{2}-4x^{3}=3-4x^{3}$$

Subtract 3 from both sides.

$$16-32x+20x^{2}-4x^{3}-3=-4x^{3}$$

Subtract 3 from 16 to get 13.

$$13-32x+20x^{2}-4x^{3}=-4x^{3}$$

Add 4x^{3} to both sides.

$$13-32x+20x^{2}-4x^{3}+4x^{3}=0$$

Combine -4x^{3} and 4x^{3} to get 0.

$$13-32x+20x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$20x^{2}-32x+13=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 20 for a, -32 for b, and 13 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-32\right)±\sqrt{\left(-32\right)^{2}-4\times 20\times 13}}{2\times 20}$$

Square -32.

$$x=\frac{-\left(-32\right)±\sqrt{1024-4\times 20\times 13}}{2\times 20}$$

Multiply -4 times 20.

$$x=\frac{-\left(-32\right)±\sqrt{1024-80\times 13}}{2\times 20}$$

Multiply -80 times 13.

$$x=\frac{-\left(-32\right)±\sqrt{1024-1040}}{2\times 20}$$

Add 1024 to -1040.

$$x=\frac{-\left(-32\right)±\sqrt{-16}}{2\times 20}$$

Take the square root of -16.

$$x=\frac{-\left(-32\right)±4i}{2\times 20}$$

The opposite of -32 is 32.

$$x=\frac{32±4i}{2\times 20}$$

Multiply 2 times 20.

$$x=\frac{32±4i}{40}$$

Now solve the equation x=\frac{32±4i}{40} when ± is plus. Add 32 to 4i.

$$x=\frac{32+4i}{40}$$

Divide 32+4i by 40.

$$x=\frac{4}{5}+\frac{1}{10}i$$

Now solve the equation x=\frac{32±4i}{40} when ± is minus. Subtract 4i from 32.

$$x=\frac{32-4i}{40}$$

Divide 32-4i by 40.

$$x=\frac{4}{5}-\frac{1}{10}i$$

The equation is now solved.

$$x=\frac{4}{5}+\frac{1}{10}i x=\frac{4}{5}-\frac{1}{10}i$$

Steps for Completing the Square

Multiply both sides of the equation by 4.

$$4\left(2-x\right)^{3}-4\left(2-x\right)^{2}=3-4x^{3}$$

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

$$4\left(8-12x+6x^{2}-x^{3}\right)-4\left(2-x\right)^{2}=3-4x^{3}$$

Use the distributive property to multiply 4 by 8-12x+6x^{2}-x^{3}.

$$32-48x+24x^{2}-4x^{3}-4\left(2-x\right)^{2}=3-4x^{3}$$

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

$$32-48x+24x^{2}-4x^{3}-4\left(4-4x+x^{2}\right)=3-4x^{3}$$

Use the distributive property to multiply -4 by 4-4x+x^{2}.

$$32-48x+24x^{2}-4x^{3}-16+16x-4x^{2}=3-4x^{3}$$

Subtract 16 from 32 to get 16.

$$16-48x+24x^{2}-4x^{3}+16x-4x^{2}=3-4x^{3}$$

Combine -48x and 16x to get -32x.

$$16-32x+24x^{2}-4x^{3}-4x^{2}=3-4x^{3}$$

Combine 24x^{2} and -4x^{2} to get 20x^{2}.

$$16-32x+20x^{2}-4x^{3}=3-4x^{3}$$

Add 4x^{3} to both sides.

$$16-32x+20x^{2}-4x^{3}+4x^{3}=3$$

Combine -4x^{3} and 4x^{3} to get 0.

$$16-32x+20x^{2}=3$$

Subtract 16 from both sides.

$$-32x+20x^{2}=3-16$$

Subtract 16 from 3 to get -13.

$$-32x+20x^{2}=-13$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$20x^{2}-32x=-13$$

Divide both sides by 20.

$$\frac{20x^{2}-32x}{20}=-\frac{13}{20}$$

Dividing by 20 undoes the multiplication by 20.

$$x^{2}+\left(-\frac{32}{20}\right)x=-\frac{13}{20}$$

Reduce the fraction \frac{-32}{20} to lowest terms by extracting and canceling out 4.

$$x^{2}-\frac{8}{5}x=-\frac{13}{20}$$

Divide -\frac{8}{5}, the coefficient of the x term, by 2 to get -\frac{4}{5}. Then add the square of -\frac{4}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{8}{5}x+\left(-\frac{4}{5}\right)^{2}=-\frac{13}{20}+\left(-\frac{4}{5}\right)^{2}$$

Square -\frac{4}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{8}{5}x+\frac{16}{25}=-\frac{13}{20}+\frac{16}{25}$$

Add -\frac{13}{20} to \frac{16}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{8}{5}x+\frac{16}{25}=-\frac{1}{100}$$

Factor x^{2}-\frac{8}{5}x+\frac{16}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{4}{5}\right)^{2}=-\frac{1}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{4}{5}\right)^{2}}=\sqrt{-\frac{1}{100}}$$

Simplify.

$$x-\frac{4}{5}=\frac{1}{10}i x-\frac{4}{5}=-\frac{1}{10}i$$

Add \frac{4}{5} to both sides of the equation.

$$x=\frac{4}{5}+\frac{1}{10}i x=\frac{4}{5}-\frac{1}{10}i$$