Solve -3(x+1)^3=24

$$- 3 ( x + 1 ) ^ { 3 } = 24$$

Solve for x (complex solution)

$$x=-3$$

$$x=\sqrt{3}i\approx 1.732050808i$$

$$x=-\sqrt{3}i\approx -0-1.732050808i$$

Solution Steps

Divide both sides by -3.

$$\left(x+1\right)^{3}=\frac{24}{-3}$$

Divide 24 by -3 to get -8.

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

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

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

Add 8 to both sides.

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

Add 1 and 8 to get 9.

$$x^{3}+3x^{2}+3x+9=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 9 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±9,±3,±1$$

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.

$$x=-3$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+3x^{2}+3x+9 by x+3 to get x^{2}+3. Solve the equation where the result equals to 0.

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

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 1 for a, 0 for b, and 3 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 1\times 3}}{2}$$

Do the calculations.

$$x=\frac{0±\sqrt{-12}}{2}$$

Solve the equation x^{2}+3=0 when ± is plus and when ± is minus.

$$x=-\sqrt{3}i x=\sqrt{3}i$$

List all found solutions.

$$x=-3 x=-\sqrt{3}i x=\sqrt{3}i$$

Solve for x

$$x=-3$$

Solution Steps

Divide both sides by -3.

$$\left(x+1\right)^{3}=\frac{24}{-3}$$

Divide 24 by -3 to get -8.

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

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

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

Add 8 to both sides.

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

Add 1 and 8 to get 9.

$$x^{3}+3x^{2}+3x+9=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 9 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±9,±3,±1$$

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.

$$x=-3$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+3x^{2}+3x+9 by x+3 to get x^{2}+3. Solve the equation where the result equals to 0.

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

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 1 for a, 0 for b, and 3 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 1\times 3}}{2}$$

Do the calculations.

$$x=\frac{0±\sqrt{-12}}{2}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$x\in \emptyset$$

List all found solutions.

$$x=-3$$