Solve s^3=729

$$s ^ { 3 } = 729$$

Solve for s

$$s=9$$

Solution Steps

Subtract 729 from both sides.

$$s^{3}-729=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -729 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±729,±243,±81,±27,±9,±3,±1$$

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.

$$s=9$$

By Factor theorem, s-k is a factor of the polynomial for each root k. Divide s^{3}-729 by s-9 to get s^{2}+9s+81. Solve the equation where the result equals to 0.

$$s^{2}+9s+81=0$$

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, 9 for b, and 81 for c in the quadratic formula.

$$s=\frac{-9±\sqrt{9^{2}-4\times 1\times 81}}{2}$$

Do the calculations.

$$s=\frac{-9±\sqrt{-243}}{2}$$

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

$$s\in \emptyset$$

List all found solutions.

$$s=9$$