If a3+b3+c3−3abc=126,a^3 + b^3 + c^3 - 3abc = 126,a3+b3+c3−3abc=126, a + b + c = 6, then the value of (ab + bc + ca) is: A8B7C5D9Answer: C. 5 Read Explanation: Solution:Given :a3+b3+c3−3abc=126a^3+b^3+c^3-3abc=126a3+b3+c3−3abc=126 and a + b + c = 6Formula used :a3+b3+c3−3abc=(a+b+c)[(a+b+c)2−3(ab+bc+ca)]a^3+b^3+c^3-3abc=(a+b+c)[(a+b+c)^2-3(ab+bc+ca)]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 Read more in App