Question

If a+b+c=6,ab+bc+ca=11,then what is a3+b3+c3-3abc

Answer 1

Answer: 18

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Work Shown:

a+b+c = 6

(a+b+c)^2 = 6^2

(a+b+c)(a+b+c) = 36

a(a+b+c)+b(a+b+c)+c(a+b+c) = 36

a^2+ab+ca+ab+b^2+ca+bc+c^2 = 36

a^2+b^2+c^2+2ab+2bc+2ca = 36

a^2+b^2+c^2+2(ab+bc+ca) = 36

a^2+b^2+c^2+2*11 = 36

a^2+b^2+c^2+22 = 36

a^2+b^2+c^2 = 36-22

a^2+b^2+c^2 = 14

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a^3+b^3+c^3-3abc = (a+b+c)*(a^2+b^2+c^2 - (ab+bc+ca) )

a^3+b^3+c^3-3abc = 6*(14 - 11)

a^3+b^3+c^3-3abc = 18

Answer 2

(a+b+c)^3=a^3 +b^3+c^3-2(ab+ac+bc)

216=a^3+b^3+c^3-2(11)

a^3+b^3+c^3=238

but ıdk how can ı find -3abc :/