Math

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Prove that for all pairs of integers a,b such that a≠b
(a²+b)(b-a)+ab(2-a-b)+(b²-a)(a+b) is a multiple of (a-b).

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For the convenience ([tex]a^{2 }[/tex]+b)(b-a) =c .

([tex]a^{2 }[/tex]+b)(b-a) +ab(2-a-b) +([tex]b^{2 }[/tex]-a)(a+b) =

=c+2ab-[tex]a^{2 }b -ab^{2 } +ab^{2 } + b^{3 } - a^{2 } -ab[/tex] =

=c+(ab-[tex]a^{2 }[/tex]) +([tex]b^{3 } - a^{2 }b[/tex]) =

=c+a(b-a) +b([tex]b^{2 } - a^{2 }[/tex]) =

=([tex]a^{2 } +b[/tex])(b-a) +a(b-a) +b(b-a)(b+a) =

=(b-a)([tex]a^{2 } +b+a+ b^{2 } +ab[/tex]) =

= - (a-b).([tex]a^{2 } +b^{2 }[/tex]+ab+a+b)