Question

If x=a(b-c), y=b(c-a), z=c(a b) then prove that (x/a) ^3+(y/b) ^3+(z/c) ^3 = 3xyz/abc?​

Answer 1

x=a(b-c) 0r x/a = (b-c)……………..(1)

y =b(c-a) or y/b = (c-a)…………………(2)

z= =c(a-b) or z/c =(a-b)…………………….(3).

Prove that:-

(x/a)^3+(y/b)^3+(z/c)^3 = 3x.y.z/a.b.c.

L.H.S.

=(x/a)^3+(y/b)^3+(z/c)^3.

We have on adding eq.(1) ,(2) & (3).

x/a+y/b+z/c=b-c+c-a+a-b =0.

If x/a+y/b+z/c=0 then

(x/a)^3+(y/b)^3+(z/c)^3=3×(x/a)×(y/b)×(z/c).

or (x/a)^3+(y/b)^3+(z/c)^3 =3.x.y.z/a.b..c.

Proved.