The sum of real values of k for which the cubic x^3-kx+k-1=0 has exactly tho distinct real solution.

Answer 1

You appear to mean "two" distinct real roots, since :tho" is just too weird to mean "one" or "three".

SO, there is a repeated root. If the two roots are p and q, with p repeated, then

(x-p)^2 (x-q) = x^3-kx+k-1
x^3 - (2p+q)x^2 + (p^2+2pq)x - p^2q = x^3-kx+k-1

if the two polynomials are the same, all the coefficients must be identical, so

2p+q = 0
p^2+2pq = -k
p^2q = k-1

q = -2p, so
p^2-4p^2 = -k
k = 3p^2
-2p^3 = k-1
1-2p^3 = k

3p^2 = 1-2p^3
2p^3+3p^2-1 = 0
p = 1/2 or -1

k = 3p^2 = 3 or 3/4

That means the cubic is

x^3 - 3x + 2
= (x-1)^2 (x+2)

or

x^3 - 3/4 x - 1/4
= (1/4)(4x^3-3x-1)
= (1/4)(x-1)(2x+1)^2