Solve (x+5)^3=27

$$( x + 5 ) ^ { 3 } = 27$$

Solve for x (complex solution)

$$x=-2$$

$$x=\frac{-13+3\sqrt{3}i}{2}\approx -6.5+2.598076211i$$

$$x=\frac{-3\sqrt{3}i-13}{2}\approx -6.5-2.598076211i$$

Solution Steps

Use binomial theorem \left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} to expand \left(x+5\right)^{3}.

$$x^{3}+15x^{2}+75x+125=27$$

Subtract 27 from both sides.

$$x^{3}+15x^{2}+75x+125-27=0$$

Subtract 27 from 125 to get 98.

$$x^{3}+15x^{2}+75x+98=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 98 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±98,±49,±14,±7,±2,±1$$

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.

$$x=-2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+15x^{2}+75x+98 by x+2 to get x^{2}+13x+49. Solve the equation where the result equals to 0.

$$x^{2}+13x+49=0$$

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, 13 for b, and 49 for c in the quadratic formula.

$$x=\frac{-13±\sqrt{13^{2}-4\times 1\times 49}}{2}$$

Do the calculations.

$$x=\frac{-13±\sqrt{-27}}{2}$$

Solve the equation x^{2}+13x+49=0 when ± is plus and when ± is minus.

$$x=\frac{-3i\sqrt{3}-13}{2} x=\frac{-13+3i\sqrt{3}}{2}$$

List all found solutions.

$$x=-2 x=\frac{-3i\sqrt{3}-13}{2} x=\frac{-13+3i\sqrt{3}}{2}$$

Solve for x

$$x=-2$$

Solution Steps

Use binomial theorem \left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} to expand \left(x+5\right)^{3}.

$$x^{3}+15x^{2}+75x+125=27$$

Subtract 27 from both sides.

$$x^{3}+15x^{2}+75x+125-27=0$$

Subtract 27 from 125 to get 98.

$$x^{3}+15x^{2}+75x+98=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 98 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±98,±49,±14,±7,±2,±1$$

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.

$$x=-2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+15x^{2}+75x+98 by x+2 to get x^{2}+13x+49. Solve the equation where the result equals to 0.

$$x^{2}+13x+49=0$$

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, 13 for b, and 49 for c in the quadratic formula.

$$x=\frac{-13±\sqrt{13^{2}-4\times 1\times 49}}{2}$$

Do the calculations.

$$x=\frac{-13±\sqrt{-27}}{2}$$

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.

$$x=-2$$