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x+y+z=3, a^2x+b^2y+c^2z=a^2+b^2+c^2,...

`x+y+z=3,
a^2x+b^2y+c^2z=a^2+b^2+c^2,
a^3x+b^3y+c^3z=a^3+b^3+c^3`
[a,b,c are unequal and ab+bc+ca `ne`=0].

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The correct Answer is:
x=y=z=1
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|{:(1,1,1),(a^2,b^2,c^2),(a^3,b^3,c^3):}|=(b-c)(c-a)(a-b)(bc+ca+ab)