Solve a^3=27

$$a ^ { 3 } = 27$$

Solve for a

$$a=3$$

Solution Steps

Subtract 27 from both sides.

$$a^{3}-27=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -27 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±27,±9,±3,±1$$

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.

$$a=3$$

By Factor theorem, a-k is a factor of the polynomial for each root k. Divide a^{3}-27 by a-3 to get a^{2}+3a+9. Solve the equation where the result equals to 0.

$$a^{2}+3a+9=0$$

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

$$a=\frac{-3±\sqrt{3^{2}-4\times 1\times 9}}{2}$$

Do the calculations.

$$a=\frac{-3±\sqrt{-27}}{2}$$

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

$$a\in \emptyset$$

List all found solutions.

$$a=3$$