If a,b,care all different and the equations ax+a2y+(a3+1)z=0,bx+b2y+(b3+1)z=0,cx+c2y+(c3+1)z=0 have a non-zero solution, then
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a
abc+1=0
b
abc−1=0
c
a2+b2+c2=ab+bc+ca
d
a3+b3+c3+3abc=0
answer is A.
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Detailed Solution
Given equations have non-zero solution⇒Δ=0⇒aa2a3+1bb2b3+1cc2c3+1=0 ⇒aa2a3bb2b3cc2c3+aa21bb21cc21=0 ⇒abc1aa21bb21cc2+1aa21bb21cc2=0 ⇒abc+1a-bb-cc-a=0 ⇒abc+1=0 ∵a, b, c are different