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If a+b+c=5 and ab+bc+ca=10, then prove that `a^(3)+b^(3)+c^(3)-3abc=-25`.

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We know that, `a^(3)+b^(3)+c^(3)-3abc=(a+b+c)(a^(2)+b^(2)+c^(2)-ab-bc-ca)`
`=(a+b+c)[a^(2)+b^(2)+c^(2)-(ab+bc+ca)]`
`=5{a^(2)+b^(2)+c^(2)-(ab+bc+ca)}=5(a^(2)+b^(2)+c^(2)-10)`
Now, a+b+c=5
Squaring both sides, we get
`(a+b+c)^(2)=5^(2)`
`implies a^(2)+b^(2)+c^(2)+2(ab+bc+ca)=25`
`therefore a^(2)+b^(2)+c^(2)+2(10)=25`
`implies a^(2)+b^(2)+c^(2)=25-20=5`
Now, `a^(3)+b^(3)+c^(3)-3abc-5(a^(2)+b^(2)+c^(2)-10)`
=5(5-10)=5(-5)=-25

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