Solve (x+8)^3=64

$$( x + 8 ) ^ { 3 } = 64$$

Solve for x (complex solution)

$$x=-4$$

$$x=-10+2\sqrt{3}i\approx -10+3.464101615i$$

$$x=-2\sqrt{3}i-10\approx -10-3.464101615i$$

Solution Steps

Use binomial theorem \left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} to expand \left(x+8\right)^{3}.

$$x^{3}+24x^{2}+192x+512=64$$

Subtract 64 from both sides.

$$x^{3}+24x^{2}+192x+512-64=0$$

Subtract 64 from 512 to get 448.

$$x^{3}+24x^{2}+192x+448=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 448 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±448,±224,±112,±64,±56,±32,±28,±16,±14,±8,±7,±4,±2,±1$$

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.

$$x=-4$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+24x^{2}+192x+448 by x+4 to get x^{2}+20x+112. Solve the equation where the result equals to 0.

$$x^{2}+20x+112=0$$

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, 20 for b, and 112 for c in the quadratic formula.

$$x=\frac{-20±\sqrt{20^{2}-4\times 1\times 112}}{2}$$

Do the calculations.

$$x=\frac{-20±\sqrt{-48}}{2}$$

Solve the equation x^{2}+20x+112=0 when ± is plus and when ± is minus.

$$x=-2i\sqrt{3}-10 x=-10+2i\sqrt{3}$$

List all found solutions.

$$x=-4 x=-2i\sqrt{3}-10 x=-10+2i\sqrt{3}$$

Solve for x

$$x=-4$$

Solution Steps

Use binomial theorem \left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} to expand \left(x+8\right)^{3}.

$$x^{3}+24x^{2}+192x+512=64$$

Subtract 64 from both sides.

$$x^{3}+24x^{2}+192x+512-64=0$$

Subtract 64 from 512 to get 448.

$$x^{3}+24x^{2}+192x+448=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 448 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±448,±224,±112,±64,±56,±32,±28,±16,±14,±8,±7,±4,±2,±1$$

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.

$$x=-4$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+24x^{2}+192x+448 by x+4 to get x^{2}+20x+112. Solve the equation where the result equals to 0.

$$x^{2}+20x+112=0$$

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, 20 for b, and 112 for c in the quadratic formula.

$$x=\frac{-20±\sqrt{20^{2}-4\times 1\times 112}}{2}$$

Do the calculations.

$$x=\frac{-20±\sqrt{-48}}{2}$$

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.

$$x=-4$$