Solve (x+3)^3=125

$$( x + 3 ) ^ { 3 } = 125$$

Solve for x (complex solution)

$$x=2$$

$$x=\frac{-11+5\sqrt{3}i}{2}\approx -5.5+4.330127019i$$

$$x=\frac{-5\sqrt{3}i-11}{2}\approx -5.5-4.330127019i$$

Solution Steps

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

$$x^{3}+9x^{2}+27x+27=125$$

Subtract 125 from both sides.

$$x^{3}+9x^{2}+27x+27-125=0$$

Subtract 125 from 27 to get -98.

$$x^{3}+9x^{2}+27x-98=0$$

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}+9x^{2}+27x-98 by x-2 to get x^{2}+11x+49. Solve the equation where the result equals to 0.

$$x^{2}+11x+49=0$$

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

$$x=\frac{-11±\sqrt{11^{2}-4\times 1\times 49}}{2}$$

Do the calculations.

$$x=\frac{-11±\sqrt{-75}}{2}$$

Solve the equation x^{2}+11x+49=0 when ± is plus and when ± is minus.

$$x=\frac{-5i\sqrt{3}-11}{2} x=\frac{-11+5i\sqrt{3}}{2}$$

List all found solutions.

$$x=2 x=\frac{-5i\sqrt{3}-11}{2} x=\frac{-11+5i\sqrt{3}}{2}$$

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+3\right)^{3}.

$$x^{3}+9x^{2}+27x+27=125$$

Subtract 125 from both sides.

$$x^{3}+9x^{2}+27x+27-125=0$$

Subtract 125 from 27 to get -98.

$$x^{3}+9x^{2}+27x-98=0$$

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}+9x^{2}+27x-98 by x-2 to get x^{2}+11x+49. Solve the equation where the result equals to 0.

$$x^{2}+11x+49=0$$

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

$$x=\frac{-11±\sqrt{11^{2}-4\times 1\times 49}}{2}$$

Do the calculations.

$$x=\frac{-11±\sqrt{-75}}{2}$$

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.

$$x=2$$