Let x+1x=1 and α,β,γ are distinct positive integers such that xa+1xα+xβ+1xβ+xγ+1xγ=0. Then minimum value of α+β+γ=

x2−x+1=0⇒x=−ω,−ω2⇒x=eiπ/3=Cisπ3⇒xα+1xα=2cos⁡απ3.  Similarly xβ+1xβ, xy+1xγ.Req⇒cos⁡απ3+cos⁡βπ3+cos⁡γπ3=0⇒12−1+12=0⇒α=1,β=3,γ=5