Question

find the value of (x-a)³+(x-b)³+(x-c) ³-3(x-a) (x-b)(x-c) where a+b+c=3x

Answer 1

hey there !!!...
..

your answer is given in attachment ..,

I think if u the pic. is not visible than see here ...

we know that,
a³+b³+c³-3abc=(a+b+c)(a²+b²+c²-ab-bc-ca)
If a+b+c=0 then,
a³+b³+c³-3abc=0×(a²+b²+c²-ab-bc-ca)=0 .....(1)

(x-a)³+(x-b)³+(x-c)³-3(x-a)(x-b)(x-c)
let a=(x-a)
b=(x-b)
and c=(x-c)
a+b+c=x-a+x-b+x-c=3x-(a+b+c)=3x-3x=0....(as a+b+c=3x)
(x-a)³+(x-b)³+(x-c)³-3(x-a)(x-b)(x-c)=0

Thus the answer is 0....

hope it helps !!!
☺☺ himanshu jha ☺☺