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 


Dividing throughout by $abc$, we get
$$ \frac{b^2+c^2 - a^2}{c^2} = \frac{a^2 + c^2 - b^2}{a^2} = \frac{a^2+b^2-c^2}{b^2}.$$
Subtracting 1 from each term, 
$$ \frac{b^2- a^2}{c^2} = \frac{ c^2 - b^2}{a^2} = \frac{a^2-c^2}{b^2}.$$
Applying Componendo et dividendo, these fractions are equal to 
$$ \frac{ b^2-a^2+c^2-b^2+a^2-c^2} { c^2 + a^2 + b^2}$$
Hence, all of them are 0, which means $a=b=c$.
