Question

Find the value of (x-a)³+(x-b)+(x-c)³-3(x-a)(x-b)(x-c) , if a+b+c=3x

Answer 1

We know that,

a³+b³+c³-3abc=(a+b+c)(a²+b²+c²-ab-bc-ca) (You must know this formula to solve this question)

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) -----(2)

(x-a)³+(x-b)³+(x-c)³-3(x-a)(x-b)(x-c)= (x-a+x-b+x-c) *((x-a)^2+(x-b)^2+(x-c)^2 -(x-a)*(x-b) -(x-b)*(x-c)-(x-c)*(x-a))

=0*((x-a)^2+(x-b)^2+(x-c)^2 -(x-a)*(x-b) -(x-b)*(x-c)-(x-c)*(x-a) ) --- From (2)

=0