Solve the equation (x + 1) (x-2) = x-4 and find: a) the sum of the squares of the roots of this equation

Solve the equation (x + 1) (x-2) = x-4 and find: a) the sum of the squares of the roots of this equation b) the difference of the cubes of the roots of this equation

Move the right side of the equation to the left side of the equation with a minus sign.
The equation will turn from
(x – 2) (x + 1) = x – 4
in
– x + 4 + (x – 2) (x + 1) = 0
Let’s open the expression in the equation and get the quadratic equation
x ^ {2} – 2 x + 2 = 0
This is an equation of the form
a * x ^ 2 + b * x + c = 0
The quadratic equation can be solved using the discriminant.
D = b ^ 2 – 4 * a * c
Because
a = 1
b = -2
c = 2
, then
D = b ^ 2 – 4 * a * c = (-2) ^ 2 – 4 * (1) * (2) = -4
Because D <0, then the equation has no real roots, but there are complex roots.
x1 = (-b + √ (D)) / (2 * a)
x2 = (-b – √ (D)) / (2 * a)
or
x_ {1} = 1 + i
x_ {2} = 1 – i
Sum of Squares – (1 + i) ^ 2 + (1 – i) ^ 2 = 0
Difference of cubes of roots (1 + i) ^ 1/3 – (1 – i) ^ 1/3 = – 2 2/3 i / 2 +2 2/3 i / {2} √ {3}