Question

If A+B+C=4, AB+BC+CA=6 find the value of A^3+B^C^3-3ABC

Answer 1

A+ B +C = 4
AB + BC+CA = 6

A^3 + B^3 +C^3 - 3ABC


as we know that
A^3 + B^3 + C^3 -3ABC = ( A+ B+ C){A^2 + B^2 + C^2 - ( AB + BC + CA)} -------(i)

let's find the value of A^2 + B^2 + C^2 first , using the identity (a + b+ c)^2

(A+B + C)^2 = {A^2 + B^2 + C^2 + 2 ( AB + BC + CA)}
substitute the given values,
(4)^2 = A^2 + B^2 + C^2 + 2 (6)
16 -12 = A^2 + B^2 + C^2

A^2 + B^2 + C^2 = 4


now,
we've got all the values needed
substitute then in (i)


( A+ B+ C){A^2 + B^2 + C^2 - ( AB + BC + CA)}
= (4) {4 -(6)}

= 4 (-2)
= -8

so,
A^3+B^3 +C^3-3ABC = -8