Solve x^3 - 3x^2 + 3x - 2 = 0.

Step-by-step solution

1. 1

   Look for the binomial-cube pattern in the coefficients. The identity

   $(x-1)^{3}=x^{3}-3x^{2}+3x-1$

   has coefficients $1,-3,3,-1$ — identical to the given cubic except in the constant term. That $1,3,3,1$ signature (Pascal's triangle) is the tell.

2. 2

   Rewrite the cubic using the pattern. Since the constant is $-2$ instead of $-1$,

   $x^{3}-3x^{2}+3x-2=\left(x^{3}-3x^{2}+3x-1\right)-1=(x-1)^{3}-1$

3. 3

   Reduce the equation to a pure cube.

   $(x-1)^{3}-1=0\ \Longrightarrow\ (x-1)^{3}=1$

4. 4

   Take the real cube root. Unlike a square root, the real cube root is single-valued: the only real number whose cube is $1$ is $1$ itself, so

   $x-1=1\ \Longrightarrow\ x=2$

5. 5

   Account for the other two roots. Over the complex numbers $(x-1)^{3}=1$ also allows $x-1=\omega$ and $x-1=\omega^{2}$ with $\omega=-\tfrac12+\tfrac{\sqrt3}{2}i$, giving the non-real roots $x=\tfrac12\pm\tfrac{\sqrt3}{2}i$. Equivalently, factoring as a difference of cubes gives $(x-2)\left(x^{2}-x+1\right)=0$, and $x^{2}-x+1$ has discriminant $-3<0$.

6. 6

   Check the real root. $2^{3}-3(2)^{2}+3(2)-2=8-12+6-2=0\ \checkmark$