Solve (x+1)^3-6(x+1)-4=0

$$( x + 1 ) ^ { 3 } - 6 ( x + 1 ) - 4 = 0$$

Solve for x

$$x=-3$$

$$x=\sqrt{3}\approx 1.732050808$$

$$x=-\sqrt{3}\approx -1.732050808$$

Solution Steps

Expand the expression.

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

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}.

$$±9,±3,±1$$

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}+3x^{2}-3x-9 by x+3 to get x^{2}-3. Solve the equation where the result equals to 0.

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

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, 0 for b, and -3 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 1\left(-3\right)}}{2}$$

Do the calculations.

$$x=\frac{0±2\sqrt{3}}{2}$$

Solve the equation x^{2}-3=0 when ± is plus and when ± is minus.

$$x=-\sqrt{3} x=\sqrt{3}$$

List all found solutions.

$$x=-3 x=-\sqrt{3} x=\sqrt{3}$$