Question

If `x/a = y/b = z/c`, prove that `"ax - by"/((a + b)(x- y)) + "by - cz"/((b + c)(y - z)) + "cz - ax"/((c + a)(z - x)` = 3

Answer 1

`x/a = y/b = z/c`
∴ x = ak, y = bk, z = ck

L.H.S. `"ax - by"/((a + b)(x- y)) + "by - cz"/((b + c)(y - z)) + "cz - ax"/((c + a)(z - x)` 

= `"a.ak - b.bk"/((a + b)(ak - bk)) + ".bk - c.ck"/((b + c)(bk - ck)) + "c.ck - a.ak"/((c + a)(ck - ak)`

= `(a^2k - b^2k)/((a + b)k(a - b)) + (b^2k^2)/((b + c)k(b - c)) + (c^2k - a^2k)/((c + a)(c - a)`

= `(k(a^2 - b^2))/(k(a^2 - b^2)) + (k(b^2 - c^2))/(k(b^2 - c^2)) + (k(c^2 - a^2))/(k(c^2 - a^2)`

= 1 + 1 + 1
= 3
= R.H.S.