Solve n^3-n=6n^2-6

$$n ^ { 3 } - n = 6 n ^ { 2 } - 6$$

Solve for n

$$n=1$$

$$n=6$$

$$n=-1$$

Solution Steps

Subtract 6n^{2} from both sides.

$$n^{3}-n-6n^{2}=-6$$

Add 6 to both sides.

$$n^{3}-n-6n^{2}+6=0$$

Rearrange the equation to put it in standard form. Place the terms in order from highest to lowest power.

$$n^{3}-6n^{2}-n+6=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 6 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±6,±3,±2,±1$$

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.

$$n=1$$

By Factor theorem, n-k is a factor of the polynomial for each root k. Divide n^{3}-6n^{2}-n+6 by n-1 to get n^{2}-5n-6. Solve the equation where the result equals to 0.

$$n^{2}-5n-6=0$$

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 -6 for c in the quadratic formula.

$$n=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 1\left(-6\right)}}{2}$$

Do the calculations.

$$n=\frac{5±7}{2}$$

Solve the equation n^{2}-5n-6=0 when ± is plus and when ± is minus.

$$n=-1 n=6$$

List all found solutions.

$$n=1 n=-1 n=6$$