How prove this is an equilateral triangle in $\Delta ABC$,$AB=c,AC=b,BC=a$and such

$$ab^2\cos{A}=bc^2\cos{B}=ca^2\cos{C}$$

show that
$\Delta ABC$ is an equilateral triangle
this problem I have solution,But not nice, and  I think this problem have more nice methods,Thank you everyone.
my solution:
$$ab^2\cdot\dfrac{b^2+c^2-a^2}{2bc}=bc^2\cdot\dfrac{a^2+c^2-b^2}{2ac}=ca^2\cdot\dfrac{a^2+b^2-c^2}{2ab}$$
$$\Longrightarrow a=b=c$$
My other idea:

$$\Longleftrightarrow\dfrac{\sin{B}}{\sin{C}}\cos{A}=\dfrac{\sin{C}}{\sin{A}}\cos{B}=\dfrac{\sin{A}}{\sin{B}}\cos{C}$$
  $$\Longleftrightarrow \sin{(2A)}\sin{B}=\sin{(2B)}\sin{C}=\sin{(2C)}\sin{A}$$
  then How prove
  $$A=B=C$$
  and have other nice methods? Thank you 


Hint:divide all the three by abc and then proceed.
