Solve x^3-x^2-x=3^3-3^2-3,w

$$x ^ { 3 } - x ^ { 2 } - x = 3 ^ { 3 } - 3 ^ { 2 } - 3 , w$$

Solve for x (complex solution)

$$x=-1-2i$$

$$x=3$$

$$x=-1+2i$$

Solution Steps

Calculate 3 to the power of 3 and get 27.

$$x^{3}-x^{2}-x=27-3^{2}-3$$

Calculate 3 to the power of 2 and get 9.

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

Subtract 9 from 27 to get 18.

$$x^{3}-x^{2}-x=18-3$$

Subtract 3 from 18 to get 15.

$$x^{3}-x^{2}-x=15$$

Subtract 15 from both sides.

$$x^{3}-x^{2}-x-15=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -15 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±15,±5,±3,±1$$

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}-x^{2}-x-15 by x-3 to get x^{2}+2x+5. Solve the equation where the result equals to 0.

$$x^{2}+2x+5=0$$

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, 2 for b, and 5 for c in the quadratic formula.

$$x=\frac{-2±\sqrt{2^{2}-4\times 1\times 5}}{2}$$

Do the calculations.

$$x=\frac{-2±\sqrt{-16}}{2}$$

Solve the equation x^{2}+2x+5=0 when ± is plus and when ± is minus.

$$x=-1-2i x=-1+2i$$

List all found solutions.

$$x=3 x=-1-2i x=-1+2i$$

Solve for x

$$x=3$$

Solution Steps

Calculate 3 to the power of 3 and get 27.

$$x^{3}-x^{2}-x=27-3^{2}-3$$

Calculate 3 to the power of 2 and get 9.

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

Subtract 9 from 27 to get 18.

$$x^{3}-x^{2}-x=18-3$$

Subtract 3 from 18 to get 15.

$$x^{3}-x^{2}-x=15$$

Subtract 15 from both sides.

$$x^{3}-x^{2}-x-15=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -15 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±15,±5,±3,±1$$

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}-x^{2}-x-15 by x-3 to get x^{2}+2x+5. Solve the equation where the result equals to 0.

$$x^{2}+2x+5=0$$

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, 2 for b, and 5 for c in the quadratic formula.

$$x=\frac{-2±\sqrt{2^{2}-4\times 1\times 5}}{2}$$

Do the calculations.

$$x=\frac{-2±\sqrt{-16}}{2}$$

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

$$x\in \emptyset$$

List all found solutions.

$$x=3$$