Math, asked by Udit1111, 1 year ago

a+b+c=9 and ab+bc+ca=26 find value of a3+b3+c3-3abc

Answers

Answered by robindav
39
use identitb:
a3 + b3 + c3  - 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)

again, (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
or a2 + b2 + c2 = (a + b + c)2 - 2ab - 2bc - 2ca

so, a3 + b3 + c3  - 3abc = (a + b + c)[(a + b + c)2– 2ab – 2bc – 2ca - ab – bc – ca]
= (a + b + c)[(a + b + c)2 – 3ab – 3bc – 3ca]
= (a + b + c)[(a + b + c)2 – 3(ab + bc + ca)]
= 9 * [92 - 3*26]
= 9 * [81 - 78]
= 9 * 3
= 27
Answered by msrajput
20
(a +b + c)2 = a2 + b2 +c2 +2(ab+bc+ca)
9^2= a2 + b2 + c2 + 2(26)
81 = a2 + b2 + c2 + 52
81 - 52 = a2 + b2 + c2
29 = a2 + b2 + c2

a3 + b3 + c3 -3abc = ( a + b + c ) (a2 + b2 + c2 -{ab +bc +ca})
a3 + b3 + c3 - 3abc = (9) (29 - 26)
a3 + b3 + c3 - 3abc = (9) (3) = 27
Similar questions