Let a, b, c ∈R be such that a+b+c>0 and abc=2. Let A=a b cb c ac a b If A2=I, then value of a3+b3+c3 is
7
2
0
-1
A2=a b cb c ac a ba b cb c ac a b=αβββαβββαwhere α=a2+b2+c2, β=bc+ca+ab.As A2=I, we geta2+b2+c2=α=1bc+ca+ab=0Now, (a+b+c)2=α+2β=1⇒ a+b+c=1We have a3+b3+c3−3abc=(a+b+c)a2+b2+c2−bc−ca−ab=1⇒a3+b3+c3=7