Solution:
Given :
a3+b3+c3−3abc=126 and a + b + c = 6
Formula used :
a3+b3+c3−3abc=(a+b+c)[(a+b+c)2−3(ab+bc+ca)]
Calculations :
126 = 6 [(6)2 - 3(ab + bc + ca)]
21 = 36 - 3(ab + bc + ca)
3(ab + bc + ca) = 15
⇒ ab + bc + ca = 5
∴ The value of ab + bc + ca is equal to 5