Question

If `x/a=y/b = z/c` show that `x^3/a^3 + y^3/b^3 + z^3/c^3 = (3xyz)/(abc)`.

Answer 1

Let `x/a=y/b = z/c` = k

=> x = ak, y = bk, z = ck

L.H.S = `x^3/a^3 + y^3/b^3 + z^3/c^3`

`= (ak)^3/(a^3) + (bk)^3/b^3 + (ck)^3/c^3`

`= (a^3k^3)/a^3 + (b^3k^3)/b^3 + (c^3k^3)/c^3`

`= k^3 + k^3 + k^3`

= `3k^3`

R.H.S = `(3xyz)/(abc)`

`= (3(ak)(bk)(ck))/(abc)`

`= (3(k^3)(abc))/(abc)`

`= 3k^3`

= L.H.S

=> L.H.S  = R.H.S

`=> x^3/a^3 + y^3/b^3 + z^3/c^3 = (3xyz)/(abc)`