Solve 0=(x-3)^3-4(x-3)^2+(x-3)+6quad0

$$0 = ( x - 3 ) ^ { 3 } - 4 ( x - 3 ) ^ { 2 } + ( x - 3 ) + 6 \quad 0$$

Solve for x

$$x=5-\sqrt{3}\approx 3.267949192$$

$$x=3$$

$$x=\sqrt{3}+5\approx 6.732050808$$

Solution Steps

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(x-3\right)^{3}.

$$0=x^{3}-9x^{2}+27x-27-4\left(x-3\right)^{2}+x-3+6\times 0$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$0=x^{3}-9x^{2}+27x-27-4\left(x^{2}-6x+9\right)+x-3+6\times 0$$

Use the distributive property to multiply -4 by x^{2}-6x+9.

$$0=x^{3}-9x^{2}+27x-27-4x^{2}+24x-36+x-3+6\times 0$$

Combine -9x^{2} and -4x^{2} to get -13x^{2}.

$$0=x^{3}-13x^{2}+27x-27+24x-36+x-3+6\times 0$$

Combine 27x and 24x to get 51x.

$$0=x^{3}-13x^{2}+51x-27-36+x-3+6\times 0$$

Subtract 36 from -27 to get -63.

$$0=x^{3}-13x^{2}+51x-63+x-3+6\times 0$$

Combine 51x and x to get 52x.

$$0=x^{3}-13x^{2}+52x-63-3+6\times 0$$

Subtract 3 from -63 to get -66.

$$0=x^{3}-13x^{2}+52x-66+6\times 0$$

Multiply 6 and 0 to get 0.

$$0=x^{3}-13x^{2}+52x-66+0$$

Add -66 and 0 to get -66.

$$0=x^{3}-13x^{2}+52x-66$$

Swap sides so that all variable terms are on the left hand side.

$$x^{3}-13x^{2}+52x-66=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -66 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±66,±33,±22,±11,±6,±3,±2,±1$$

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.

$$x=3$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-13x^{2}+52x-66 by x-3 to get x^{2}-10x+22. Solve the equation where the result equals to 0.

$$x^{2}-10x+22=0$$

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, -10 for b, and 22 for c in the quadratic formula.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 1\times 22}}{2}$$

Do the calculations.

$$x=\frac{10±2\sqrt{3}}{2}$$

Solve the equation x^{2}-10x+22=0 when ± is plus and when ± is minus.

$$x=5-\sqrt{3} x=\sqrt{3}+5$$

List all found solutions.

$$x=3 x=5-\sqrt{3} x=\sqrt{3}+5$$