if a+b+c = 9 and ab+bc+ca = 23 then a cube + b cube + c cube- 3abc=?
a3+b3+c3- 3abc = (a+b+c) (a2+b2+c2-ab-bc-ca)
a+b+c= 9 (given)
ab+ bc + ca = 23 (given)
So, a3+b3+c3- 3abc = 9 (a2+b2+c2- 23) … (1)
(a+b+c)2= a2+b2+c2+ 2ab+ 2bc+ 2ca
92= a2+b2+c2+ 2(ab+bc+ca)
81= a2+b2+c2+ 2(23)
81 = a2+b2+c2+ 46
a2+b2+c2= 35
From (1), we get,
a3+b3+c3- 3abc = 9 (35-23) = 108