Solve y=(x-1)^3-(x-2)^3

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

Solve for y

$$y=3x^{2}-9x+7$$

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}.

$$y=x^{3}-3x^{2}+3x-1-\left(x-2\right)^{3}$$

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

$$y=x^{3}-3x^{2}+3x-1-\left(x^{3}-6x^{2}+12x-8\right)$$

To find the opposite of x^{3}-6x^{2}+12x-8, find the opposite of each term.

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

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

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

Combine -3x^{2} and 6x^{2} to get 3x^{2}.

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

Combine 3x and -12x to get -9x.

$$y=3x^{2}-9x-1+8$$

Add -1 and 8 to get 7.

$$y=3x^{2}-9x+7$$

Solve for x (complex solution)

$$x=-\frac{\sqrt{12y-3}}{6}+\frac{3}{2}$$

$$x=\frac{\sqrt{12y-3}}{6}+\frac{3}{2}$$

Solve for x

$$x=-\frac{\sqrt{12y-3}}{6}+\frac{3}{2}$$

$$x=\frac{\sqrt{12y-3}}{6}+\frac{3}{2}\text{, }y\geq \frac{1}{4}$$