The system of equations x−cy−bz=0,cx−y+az=0 and bx+ay−z=0 have non-trivial solution if 1cb−c−1a−ba−1=0⇒1(1−a2)+c(−c−ab)−b(ca+b)=0 ⇒a2+b2+c2+2abc=1.
Let a,b,c be any real numbers. Suppose that there are real numbers x,y,z not all zero such that x= cy+bz,y=az+cx and z=bx+ay. Then a2+b2+c2+2abc is equal to
Held on 30 Apr 2008 · Verified 6 Jul 2026.
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