1066392 1066392

15-06-2020
Mathematics

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if a+b+c=16 and a^2+b^2+c^2=90 then find the value a^3+b^3+c^3-3abc

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09pqr4sT 09pqr4sT

15-06-2020

[tex](a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\\(16)^2=90+2(ab+bc+ca)\\256-90=2(ab+bc+ca)\\166/2=ab+bc+ca\\83=ab+bc+ca\\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))\\16(90-(83))\\16(7)\\=112[/tex]

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