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n^{3}-n-9n^{2}=-9
Subtract 9n^{2} from both sides.
n^{3}-n-9n^{2}+9=0
Add 9 to both sides.
n^{3}-9n^{2}-n+9=0
Rearrange the equation to put it in standard form. Place the terms in order from highest to lowest power.
±9,±3,±1
By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 9 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.
n=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^{2}-8n-9=0
By Factor theorem, n-k is a factor of the polynomial for each root k. Divide n^{3}-9n^{2}-n+9 by n-1 to get n^{2}-8n-9. Solve the equation where the result equals to 0.
n=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 1\left(-9\right)}}{2}
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, -8 for b, and -9 for c in the quadratic formula.
n=\frac{8±10}{2}
Do the calculations.
n=-1 n=9
Solve the equation n^{2}-8n-9=0 when ± is plus and when ± is minus.
n=1 n=-1 n=9
List all found solutions.