Question

If a+b+c=9 and AB+BC+CA=18,then what is the value of a^3+b^3+c^3-3abc???​

Answer 1

a+b+c=9

AB+BC+CA=18

By using this identity:

(a+b+c)²=a²+b²+c²+2(AB+BC+CA)

(9)²=a²+b²+c² +2(18)

81-36=a²+b²+c²

45=a²+b²+c²

Now,using this identity:

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

(9)(45-18)

=9×27=243