Solve r^3=125

$$r ^ { 3 } = 125$$

Solve for r

$$r=5$$

Solution Steps

Subtract 125 from both sides.

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

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.

$$r=5$$

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

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

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.

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

Do the calculations.

$$r=\frac{-5±\sqrt{-75}}{2}$$

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

$$r\in \emptyset$$

List all found solutions.

$$r=5$$