Question 252215
Let's look at the first equation set. 

x^2 -2X+1 = -X^3+3x^2 - 3X + 3
X^3 -2X^2 + X -2 = 0
x^2(X-2) + 1(X-2) = 0
(X^2+1)(X-2) = 0
So, we set each part = 0.
X^2 + 1 = 0
X^2 = -1. Here we have either no solution or +-i
X-2 = 0.
X=2. This is the only real solution.

Now, let's look at the second equation set.
X^4 - 5X^2 +4 = 2
X^4 - 5X^2 +2 = 0
Let U = X^2. This a "dummy variable" substitution. We get
U^2 - 5U +2 = 0.
We apply the quadratic to get
U = (5 +- sqrt(25-8))/2
U = (5 +- sqrt(17))/2.
We resubstitute X^2 for U and get
X^2 = (5 +- sqrt(17))/2.
X = +-sqrt((5 +-sqrt(17))/2). These are both real solutions.