Question

if alpha, beta and gamma are zeroes of 4x^3-32x^2+25x+21=0, then value of
[alpha^3/{(alpha-beta)(alpha-gamma)}]
+
[beta^3/{(beta-alpha)(beta-gamma)}]
+
[gamma^3/{(gamma-alpha)(gamma-beta)}]
is?

Answer 1

Looking at the 21, I suspect that roots could be ±1, ±3, ±7
after a few tries, when x = 7
4(7)^3-32(7)x^2+25(7)x+21=0

So I did synthetic division and
4x^3-32x^2+25x+21=0
(x-7)(4x^2 - 4x - 3) = 0
(x-7)(2x + 1)(2x - 3) = 0

so your zeros are 7, -1/2 and 3/2

Now perform that gibberish on them that you stated as:

[alpha^3/{(alpha-beta)(alpha-gamma)}]
+
[beta^3/{(beta-alpha)(beta-gamma)}]
+
[gamma^3/{(gamma-alpha)(gamma-beta)}]