x4 + Px3 + Qx2 + Rx + S = 0
Let a, b, c, d be the roots. It is given that sum of two roots is equal to sum of other two roots.
So a + b = c + d ...(1)
We know that a + b + c + d = – P

a + b + a + b = – P
i.e. a + b = – P/2 = c + d ….......(2)
Also, ab + ac +ad + bc + bd + cd = Q

ab + a(c + d) + b(c + d) + cd = Q
ab + (a+b)(c+d) + cd = Q

ab + (– P/2)
2 + cd = Q [From (1) & (2)]
i.e. ab + cd = Q – P2 / 4 …..........(3)
abc + abd + acd + bcd = – R

ab(c + d) + cd(a + b) = – R
ab(a + b) + cd(a + b) = (a + b)(ab + cd) = – R [From (1)]
(– P/2)(Q – P2/4) = – R [From (2) & (3)]
– (PQ/2) + (P
3/8) = – R

– 4PQ + P
3 = – 8R
P3 + 8R = 4PQ