Question

(a2-63)3 +(62-c2,3 +(c2-a2)3
(a-b)3 +(b-c)3 +(c-a)3
= (a + b)(b + c)(c + a)​

Answer 1

Answer:

L.H.S

by using a2-b2=(a+b) (a-b)

[(a+b)(a-b)]3+[(b+c)(b-c)]3+[(c+a)(c-a)]3

now let a=[(a+b)(a-b)]3,b=[(b+c)(b-c)]3,c=[(c+a)(c-a)]3

by adding a,b,&c we get

=[(a+b)(a-b)]+[(b+c)(b-c)]+[(c+a)(c-a)]

=a2-ab+ab-b2+b2-bc+bc-c2+c2+ac-ac-a2

=o

if a+b+c=o then a3+b3+c3=3abc

=>3[(a+b)(a-b)(b+c)(b-c)(c+a)(c-a)]

=>o

=>L.H.S=R.H.S

hence proved