if u find (which i did sum wierd way)
∑α
∑αβ
∑αβγ
u could do this
x^3 - (∑α)x^2 + (∑αβ)x - (∑αβγ) = 0
then:
∑α = 4
∑αβ = 3
∑αβγ = -2
so
x^3 - 4x^2 + 3x - (-2) = 0
and since alpha satisfies the eqn
α^3 - 4α^2 + 3α +2 = 0
now times that thru alpha to get:
α^4 - 4α^3 + 3α^2 +2α = 0
α^4 = 4α^3 - 3α^2 - 2α
now do that with β and γ, then add the eqns and u get:
∑(α^4) = 4[∑(α^3)] - 3[∑(α^2)] - 2α
substitute your numbers from the answer
∑(α^4) = 4[22] - 3[10] - 2[4]
∑(α^4) = 88 - 30 - 8
∑(α^4) = 50
probly not the best way to do it in an exam.. i actually missed out cuz i had a trial for economics.. which was just as sh*t.