<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] $$(a-c)(a+c)^2+bc(a+c)=ab^2$$
$$\implies a^3-a b^2-a c^2+b c^2+a^2 c+a b c-c^3=0$$
$$\implies a(a^2-b^2)-c^2(a-b)+ca(a+b)-c^3=0$$
$$\implies a(a+b)(a-b)-c^2(a-b)+ca(a+b)-c^3=0$$
$$\implies (a-b)(a(a+b)-c^2)+c(a(a+b)-c^2)=0$$
$$\implies (a-b+c)(a(a+b)-c^2)=0$$
Dividing the both sides by $a-b+c\gt 0$ gives
$$c^2=a^2+ab$$
from which we have $c\gt a$.
Case 1 : $c^2=a^2+b^2$
$a=b\implies c=\sqrt 2\ a$, which contradicts that the triangle has $48^\circ$.
Case 2 : $c^2\gt a^2+b^2$.
$a\gt b\implies C\gt 90^\circ, A=48^\circ, B\lt 42^\circ$. 
By the law of cosines,
$$a^2=b^2+a^2+ab-2b\sqrt{a^2+ab}\ \cos(48^\circ)\implies \frac ba=4\cos^2(48^\circ)-1$$
Also,
$$c^2=a^2+ab=a^2+b^2-2ab\cos C\implies \cos C=\frac{b}{2a}-\frac 12=2\cos^2(48^\circ)-1=\cos(96^\circ)$$
Therefore, $A=48^\circ,B=36^\circ,C=96^\circ$.
Case 3 : $c^2\lt a^2+b^2$.
$a\lt b$ and $C\lt 90^\circ$.
If $B=48^\circ$, then
$$\begin{align}&b^2=a^2+a^2+ab-2a\sqrt{a^2+ab}\ \cos(48^\circ)\\&\implies 2\cos(48^\circ)=\sqrt{2\cos C+2}\ (1-2\cos C)\\&\implies 2\cos(48^\circ)=2\cos\frac{C}{2}\left(1-2\left(2\cos^2\frac{C}{2}-1\right)\right)\\&\implies\cos(48^\circ)=3\cos\frac C2-4\cos^3\frac C2\\&\implies\cos(48^\circ)=-\cos\left(\frac{3C}{2}\right)\\&\implies C=32^\circ,88^\circ\end{align}$$ 
If $C=48^\circ$, then
$$c^2=a^2+ab=a^2+b^2-2ab\cos(48^\circ)\implies \cos(48^\circ)=\frac 12(4\cos^2A-1)-\frac 12=\cos(2A)$$$$\implies A=24^\circ$$
Therefore, the answer is
$$\color{red}{(A,B,C)=(48^\circ, 36^\circ, 96^\circ),(44^\circ,48^\circ, 88^\circ),(24^\circ,108^\circ,48^\circ)}$$
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