Solve y^3=125

$$y ^ { 3 } = 125$$

Solve for y (complex solution)

$$y=\frac{-5\sqrt{3}i-5}{2}\approx -2.5-4.330127019i$$

$$y=5$$

$$y=\frac{-5+5\sqrt{3}i}{2}\approx -2.5+4.330127019i$$

Solution Steps

Subtract 125 from both sides.

$$y^{3}-125=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -125 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±125,±25,±5,±1$$

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.

$$y=5$$

By Factor theorem, y-k is a factor of the polynomial for each root k. Divide y^{3}-125 by y-5 to get y^{2}+5y+25. Solve the equation where the result equals to 0.

$$y^{2}+5y+25=0$$

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, 5 for b, and 25 for c in the quadratic formula.

$$y=\frac{-5±\sqrt{5^{2}-4\times 1\times 25}}{2}$$

Do the calculations.

$$y=\frac{-5±\sqrt{-75}}{2}$$

Solve the equation y^{2}+5y+25=0 when ± is plus and when ± is minus.

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

List all found solutions.

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

Solve for y

$$y=5$$

Solution Steps

Subtract 125 from both sides.

$$y^{3}-125=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -125 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±125,±25,±5,±1$$

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.

$$y=5$$

By Factor theorem, y-k is a factor of the polynomial for each root k. Divide y^{3}-125 by y-5 to get y^{2}+5y+25. Solve the equation where the result equals to 0.

$$y^{2}+5y+25=0$$

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, 5 for b, and 25 for c in the quadratic formula.

$$y=\frac{-5±\sqrt{5^{2}-4\times 1\times 25}}{2}$$

Do the calculations.

$$y=\frac{-5±\sqrt{-75}}{2}$$

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

$$y\in \emptyset$$

List all found solutions.

$$y=5$$