<s>[INST] If the sides of a triangle satisfy $(a-c)(a+c)^2+bc(a+c)=ab^2$, and if one angle is $48^\circ$, then find the other angles. 
In triangle $ABC$ one angle of which is $48^{\circ}$, length of the sides satisfy the equality:
  $$(a-c)(a+c)^2+bc(a+c)=ab^2$$
  Find the value in degrees the other two angles of the triangle.

I have no idea how to solve this problem
 [/INST] From  $c^2-a^2=ab,$
using $(i)$Sine Law,
$(ii)$Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $,  
and $(iii)\sin(C+A)=\cdots=\sin B$
$$\sin B\sin(C-A)=\sin A\sin B$$
As $\sin B\ne0,\implies\sin(C-A)=\sin A$
$\implies C-A=n\pi+(-1)^nA$ where $n$ is any integer 
If $n$ is odd, $n=2m+1$(say) $\implies C-A=(2m+1)\pi+(-1)A$
$\iff C=(2m+1)\pi$ which is impossible as $0<C<\pi$
If $n$ is even, $n=2m$(say) $\implies C-A=2m\pi+A\iff C=2m\pi+2A$
As $0<C,A<\pi; C=2A$
We also have $A+B+C=\pi$
Now check for the three cases namely, $A=48^\circ, B=48^\circ,C=48^\circ,$
</s>