Solve (x+2)^3=-64

$$( x + 2 ) ^ { 3 } = - 64$$

Solve for x (complex solution)

$$x=-6$$

$$x=2\sqrt{3}i\approx 3.464101615i$$

$$x=-2\sqrt{3}i\approx -0-3.464101615i$$

Solution Steps

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

$$x^{3}+6x^{2}+12x+8=-64$$

Add 64 to both sides.

$$x^{3}+6x^{2}+12x+8+64=0$$

Add 8 and 64 to get 72.

$$x^{3}+6x^{2}+12x+72=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 72 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±72,±36,±24,±18,±12,±9,±8,±6,±4,±3,±2,±1$$

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.

$$x=-6$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+6x^{2}+12x+72 by x+6 to get x^{2}+12. Solve the equation where the result equals to 0.

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

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 12 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 1\times 12}}{2}$$

Do the calculations.

$$x=\frac{0±\sqrt{-48}}{2}$$

Solve the equation x^{2}+12=0 when ± is plus and when ± is minus.

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

List all found solutions.

$$x=-6 x=-2i\sqrt{3} x=2i\sqrt{3}$$

Solve for x

$$x=-6$$

Solution Steps

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

$$x^{3}+6x^{2}+12x+8=-64$$

Add 64 to both sides.

$$x^{3}+6x^{2}+12x+8+64=0$$

Add 8 and 64 to get 72.

$$x^{3}+6x^{2}+12x+72=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 72 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±72,±36,±24,±18,±12,±9,±8,±6,±4,±3,±2,±1$$

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.

$$x=-6$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+6x^{2}+12x+72 by x+6 to get x^{2}+12. Solve the equation where the result equals to 0.

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

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 12 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 1\times 12}}{2}$$

Do the calculations.

$$x=\frac{0±\sqrt{-48}}{2}$$

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.

$$x=-6$$