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If a,b,c are all different and the equations $a x + a^{2} y + (a^{3} + 1) z = 0, b x + b^{2} y + (b^{3} + 1) z = 0$ , $c x + c^{2} y + (c^{3} + 1) z = 0$ have a nonzero solution, then

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abc+1=0

abc−1=0

a2+b2+c3=ab+bc+ca

a3+b3+c3+3abc=0

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detailed solution

Correct option is A

Given equation have non zero solution ⇒aa2a3+1bb2b3+1cc2c3+1=0⇒aa2abb2b3cc2c3+aa21bb21cc21=0⇒abc1aa21bb21cc2+1aa21bb21cc2=0⇒(abc+1)1aa21bb21cc2=0⇒abc=−1

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If a,b,c are all different and the equations ax+a2y+a3+1z=0, bx+b2y+b3+1z=0 , cx+c2y+c3+1z=0 have a nonzero solution, then

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free study material

If a,b,c are all different and the equations $a x + a^{2} y + (a^{3} + 1) z = 0, b x + b^{2} y + (b^{3} + 1) z = 0$ , $c x + c^{2} y + (c^{3} + 1) z = 0$ have a nonzero solution, then