Question

A + b + c = 11 ab + bc + ac = 17 , then a³+ b³+ c³- 3abc = ?

Answer 1

Given:
a+b+c=11
ab + bc + ac = 17

(a+b+c)²= a²+b²+c²+2(ab+bc+ca)

(11)² = a²+b²+c²+2(17)
121= a²+b²+c²+34
121 -34 = a²+b²+c²
87 =a²+b²+c²

a³+b³+c³-3abc= (a+b+c)(a²+b²+c²- ab-bc-ca)
a³+b³+c³-3abc= (a+b+c)(a²+b²+c²- ((ab+bc+ca))

a³+b³+c³-3abc= (11) ((87)- (17))

a³+b³+c³-3abc= 11 × 70


a³+b³+c³-3abc= 770

================================
Hope this will help you....

Answer 2

Given
a+b+c=11
ab+bc+ac=17

a³+b³+c³-3abc = (a+b+c)(a²+b²+c²-ab-bc-ca)
(a+b+c)²= a²+b²+c²+2ab+2bc+2ca
a²+b²+c²=(a+b+c)²-2ab-2bc-2ca
a²+b²+c²=(a+b+c)²-2(ab+bc+ca)
=(11)^2-2 (17)
=121-34
=87 -------equation (i)
Now,
a³+b³+c³-3abc = (a+b+c)(a²+b²+c²-ab-bc-ca)

=(a+b+c)(a²+b²+c²-(ab+bc+ca))
=11*(87-17)
=11 (70)
=770