if a+b+c=6 and a²+b²+c²=14 find a²+b²+c² - 3abc
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Given a+b+c=6
a^2+b^2+c^2=14
We know that
a^3+b^3+c^3-3abc=(a+b+c) (a^2+b^2+c^2-ab-bc-ac).... Eq1
(a+b+c) ^2=a^2+b^2+c^2+2ab+2bc+2ca
(6)^2=14+2ab+2bc+2ca
36-14=2(ab+bc+ca)
(ab+bc+ca)=22/2
ab+bc+ca=11
Eq1....
a^3+b^3+c^3-3abc=(6){14-(11)}
a^3+b^3+c^3-3abc=6(3)
a^3+b^3+c^3-3abc=18
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