If a + b + c = 9, ab + bc + ca = 26, a3 + b3 = 91, b3 + c3 = 72 and c3 + a3 = 35, then what is the value of abc?
Explanation:
Using algebraic identities,
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
By putting the respective values given in question,
⇒ (9)2 = a2 + b2 + c2 + 2(ab + bc + ca) [? ab + bc + ca = 26]
⇒ (9)2 = a2 + b2 + c2 + 2(26)
⇒ a2 + b2 + c2 = 81 - 52 = 29
Given equations,
a3 + b3 = 91 ----(1)
b3 + c3 = 72 ----(2)
c3 + a3 = 35 ----(3)
On adding (1), (2) and (3)
a3 + b3 + b3 +c2 + c2 + a3 = 91 + 72 + 35
⇒ 2(a3 + b3 + c3) = 198
⇒ a3 + b3 + c3 = 99
Using algebraic identities,
a3 + b3 + c3 - 3abc = (a + b + c) (a2 + b2 + c2 - ab - bc - ca)
By putting the respective values,
⇒ 99 - 3abc = 9 (29 - 26) [? ab + bc + ca = 26 and a + b + c = 9]
⇒ 3abc = 99 - 27
⇒ abc = 72/3
∴ abc = 24
By: MIRZA SADDAM HUSSAIN ProfileResourcesReport error