Math miscellaneousHard

Question

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

Options

A.2
B.-1
C.0
D.1

Solution

The system of equations x - cy - bz = 0, cx - y + az = 0 and bx + ay - z = 0 have non-trivial solution if
= 0 ⇒ 1(1 - a2) + c(-c - ab) - b(ca + b) = 0
⇒     a2 + b2 + c2 + 2abc = 1.

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