Question

a+b+c=15 and a square + b square + c square =83

Find the value of a cube + b cube + c cubeb - 3abc

Answer 1

Given, a + b + c = 15 and a2 + b2 + c2 = 83

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

Again, (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

⇒ 2(ab + bc + ca) = (a + b + c)2 - (a2 + b2 + c2)

⇒ 2(ab + bc + ca) = (15)2 - 83 = 225 - 83 = 142

⇒ ab + bc + ca = 71

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

⇒a3 + b3 + c3 - 3abc = (15) [83 - (71)]

⇒a3 + b3 + c3 - 3abc = (15) [12] = 180