Tuesday, 26 January 2016

If α+β+γ=1, α^2+β^2+γ^2=6, α^3+β^3+γ^3=8 then α^4+β^4+γ^4=?

If α+β+γ=1, α2+β2+γ2=6, α3+β3+γ3=8 then α4+β4+γ4=?
Solution: If x, y and z replaces α β and γ for easy writing, then
(X2+y2+z2)=x4+y4+z4+2{(xy)2+(yz)2+(zx)2}   -----(1)
Therefore, solving with given values it goes like x4+y4+z4=36-2{(xy)2+(yz)2+(zx)2}   -(2)
(X+Y+Z)2=X2+y2+z2+2XY+2YZ+2ZX-------(3)
Therefore, solving with given values it goes like (xy)2+(yz)2+(zx)2=1-4xyz-----(4)
X3+y3+z3-3xyz=(x+y+z) (X2+y2+z2-XY-YZ-ZX) -----(5)
Therefore, solving with given values it goes like xyz=-2---------(7)
Then using (7) in (4 ) and followed by solving 2 by using (4) , it goes like  x4+y4+z4=18 (Answer).




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