Solve u^3=216,u

$$u ^ { 3 } = 216 , u$$

Solve for u

$$u=6$$

Solution Steps

Subtract 216 from both sides.

$$u^{3}-216=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -216 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±216,±108,±72,±54,±36,±27,±24,±18,±12,±9,±8,±6,±4,±3,±2,±1$$

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.

$$u=6$$

By Factor theorem, u-k is a factor of the polynomial for each root k. Divide u^{3}-216 by u-6 to get u^{2}+6u+36. Solve the equation where the result equals to 0.

$$u^{2}+6u+36=0$$

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, 6 for b, and 36 for c in the quadratic formula.

$$u=\frac{-6±\sqrt{6^{2}-4\times 1\times 36}}{2}$$

Do the calculations.

$$u=\frac{-6±\sqrt{-108}}{2}$$

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

$$u\in \emptyset$$

List all found solutions.

$$u=6$$