If $$x=a(b-c),y=b(c-a),z=c(a-b)$$Find $$(frac{x}{a})^3+(frac{y}{b})^3+(frac{z}{c})$$

Options:

$$text{1: }frac{xyz}{abc}$$

$$text{2: }frac{xyz}{3abc}$$

$$text{3: }frac{abc}{xyz}$$

$$text{4: }frac{3abc}{xyz}$$

What I tried:

Try 1.

$$(frac{a(b-c)}{a})^3+(frac{b(c-a)}{b})^3+(frac{c(a-b)}{c}$$

But that leaves us with

$$(b-c)^3+(c-a)^3+(a-b)^3$$

That doesn’t match with any options given above.

Try 2

Since

$$frac{x}{(b-c)}=a,frac{y}{(c-a)})=b,frac{z}{(a-b)}=c$$

$$(frac{a(b-c)(b-c)}{x})^3+(frac{b(c-a)(c-a)}{y})^3+(frac{c(a-b)(a-b)}{z}$$

But that leaves us with:

$$(frac{acdot (b^2-2bc+c^2)}{x})^3+(frac{bcdot (c^2-2ac+a^2)}{y})^3+(frac{ccdot (a^2-2ab+b^2)}{z}$$

So is there any way where I can get the correct answer?

I am open to any advice and answers

If I have missed something obvious, please be gentle.