Solve (x+3)^3-(x-1)^3=0

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

Solve for x (complex solution)

$$x=\frac{2\sqrt{3}i}{3}-1\approx -1+1.154700538i$$

$$x=-\frac{2\sqrt{3}i}{3}-1\approx -1-1.154700538i$$

Steps Using the Quadratic Formula

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

$$x^{3}+9x^{2}+27x+27-\left(x-1\right)^{3}=0$$

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

$$x^{3}+9x^{2}+27x+27-\left(x^{3}-3x^{2}+3x-1\right)=0$$

To find the opposite of x^{3}-3x^{2}+3x-1, find the opposite of each term.

$$x^{3}+9x^{2}+27x+27-x^{3}+3x^{2}-3x+1=0$$

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

$$9x^{2}+27x+27+3x^{2}-3x+1=0$$

Combine 9x^{2} and 3x^{2} to get 12x^{2}.

$$12x^{2}+27x+27-3x+1=0$$

Combine 27x and -3x to get 24x.

$$12x^{2}+24x+27+1=0$$

Add 27 and 1 to get 28.

$$12x^{2}+24x+28=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 12 for a, 24 for b, and 28 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-24±\sqrt{24^{2}-4\times 12\times 28}}{2\times 12}$$

Square 24.

$$x=\frac{-24±\sqrt{576-4\times 12\times 28}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-24±\sqrt{576-48\times 28}}{2\times 12}$$

Multiply -48 times 28.

$$x=\frac{-24±\sqrt{576-1344}}{2\times 12}$$

Add 576 to -1344.

$$x=\frac{-24±\sqrt{-768}}{2\times 12}$$

Take the square root of -768.

$$x=\frac{-24±16\sqrt{3}i}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{-24±16\sqrt{3}i}{24}$$

Now solve the equation x=\frac{-24±16\sqrt{3}i}{24} when ± is plus. Add -24 to 16i\sqrt{3}.

$$x=\frac{-24+16\sqrt{3}i}{24}$$

Divide -24+16i\sqrt{3} by 24.

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

Now solve the equation x=\frac{-24±16\sqrt{3}i}{24} when ± is minus. Subtract 16i\sqrt{3} from -24.

$$x=\frac{-16\sqrt{3}i-24}{24}$$

Divide -24-16i\sqrt{3} by 24.

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

The equation is now solved.

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

Steps for Completing the Square

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

$$x^{3}+9x^{2}+27x+27-\left(x-1\right)^{3}=0$$

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

$$x^{3}+9x^{2}+27x+27-\left(x^{3}-3x^{2}+3x-1\right)=0$$

To find the opposite of x^{3}-3x^{2}+3x-1, find the opposite of each term.

$$x^{3}+9x^{2}+27x+27-x^{3}+3x^{2}-3x+1=0$$

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

$$9x^{2}+27x+27+3x^{2}-3x+1=0$$

Combine 9x^{2} and 3x^{2} to get 12x^{2}.

$$12x^{2}+27x+27-3x+1=0$$

Combine 27x and -3x to get 24x.

$$12x^{2}+24x+27+1=0$$

Add 27 and 1 to get 28.

$$12x^{2}+24x+28=0$$

Subtract 28 from both sides. Anything subtracted from zero gives its negation.

$$12x^{2}+24x=-28$$

Divide both sides by 12.

$$\frac{12x^{2}+24x}{12}=-\frac{28}{12}$$

Dividing by 12 undoes the multiplication by 12.

$$x^{2}+\frac{24}{12}x=-\frac{28}{12}$$

Divide 24 by 12.

$$x^{2}+2x=-\frac{28}{12}$$

Reduce the fraction \frac{-28}{12} to lowest terms by extracting and canceling out 4.

$$x^{2}+2x=-\frac{7}{3}$$

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+2x+1^{2}=-\frac{7}{3}+1^{2}$$

Square 1.

$$x^{2}+2x+1=-\frac{7}{3}+1$$

Add -\frac{7}{3} to 1.

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

Factor x^{2}+2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+1\right)^{2}=-\frac{4}{3}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{-\frac{4}{3}}$$

Simplify.

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

Subtract 1 from both sides of the equation.

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