A relation between sides in triangle $ABC$

Question

If we have below relation in triangle $ABC$ what is $\hat{B}$ $$60\\120\\135\\150$$ the relation is $$a^2+c^2+3a^2c=(a+c)(ac+b^2)-3ac^2$$ I tried it many times ,but I get stuck ...may I use cosine theorem ?

Can someone give a ray of hope? please ... Thanks in advance.

Remark: It was given from one my colleague. I am in doubt with it, because if it was $$a^3+c^3+3a^2c=(a+c)(ac+b^2)-3ac^2$$ I can solve it easily, and $\hat{B}=120$ but in this case, I don't know what to do.

Answer 1

By the Law of cosines, $$b^2=a^2+c^2-2ac\cos(\hat{B}).$$ Hence, by the relation it follows that $$\frac{a^2+c^2+3a^2c-3ac^2}{(a+c)}-ac=b^2=a^2+c^2-2ac\cos(\hat{B})$$ which implies that $$\cos(\hat{B})=\frac{(a+c)(a-c)^2-a^2-c^2}{2ac(a+c)}.$$ This means that $\cos(\hat{B})$ depends on the values of $a$ and $c$.

P.S. Note that with the relation $a^3+c^3+3a^2c=(a+c)(ac+b^2)-3ac^2$, following the same approach we get $\cos(\hat{B})=-1/2$, that is $B=120^{\circ}$.

Answer 2

So we can rewrite the formula like this $$a^2+ac+c^2 =b^2$$ so we have $$a^2+ac+c^2 =b^2 =a^2+c^2-2ac\cos \beta$$ and thus we get $$ \cos \beta =-{1\over 2}\Longrightarrow \beta = 120^{\circ}$$