Solve x^3-9x^2+27x-27=0

$$x ^ { 3 } - 9 x ^ { 2 } + 27 x - 27 = 0$$

Solve for x

$$x=3$$

Solution Steps

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.

$$x=3$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-9x^{2}+27x-27 by x-3 to get x^{2}-6x+9. Solve the equation where the result equals to 0.

$$x^{2}-6x+9=0$$

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, -6 for b, and 9 for c in the quadratic formula.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 1\times 9}}{2}$$

Do the calculations.

$$x=\frac{6±0}{2}$$

Solutions are the same.

$$x=3$$