In $\triangle ABC$, if $b^2 = a(c+a)$ and $c^2 = b(a+b)$, then $\cos A\cos B\cos C = -\frac{1}{8}$

Question

In a $\triangle ABC$ with $$b^2 = a(c+a) \quad\text{and}\quad c^2 = b(a+b)$$ prove that $$\cos A\cos B\cos C = -\frac{1}{8}$$

I am stuck in a step where I took the relation $$\cos A = \frac{b^2+c^2-a^2}{2bc}$$ modified with the given relation of $b^2 = a(c+a)$ and got $\cos A = \frac{a+c}{2b}$. Similarly, I found out for $\cos B$. However, $\cos C$ is coming up with complicated relation which is not possibly anywhere close to answer.

Answer 1

$$b^2=ac+a^2 \implies b^2-a^2=ac$$

$$c^2=ab+b^2 \implies c^2-b^2=ab$$

\begin{align}\cos A \cos B \cos C &= \frac{b^2+c^2-a^2}{2bc}\frac{a^2+c^2-b^2}{2ac}\frac{a^2+b^2-c^2}{2ab} \\&= \frac{ac+c^2}{2bc}\frac{a^2+ab}{2ac}\frac{a^2-ab}{2ab} \\&=\frac{a^2c(a+c)(a+b)(a-b)}{8(abc)^2}\\ &=\frac{(a+c)(a^2-b^2)}{8b^2c}\\ &=-\frac{(a+c)ac}{8b^2c}\end{align}

Can you take it from here?

Answer 2

Using Law of Sines on $$ca=b^2-a^2$$

$$\sin C\sin A=\sin^2B-\sin^2A=\sin(B+A)\sin(B-A)$$ using Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $

Again as $B+A=\pi-C,\sin(B+A)=\cdots=\sin C$

So, we have $$\sin C\sin A=\sin C\sin(B-A)$$

As $0<C<\pi,\sin C>0,$ hence $\ne0$ $$\implies \sin A=\sin(B-A)$$

$\implies $ either $A\equiv B-A\pmod{2\pi}\iff B\equiv2A$

As $0<A,B<\pi, B=2A$

Or $A=\pi-(B-A)\pmod{2\pi}\iff B\equiv\pi$ which is impossible

Similarly, $c^2=b(a+b)\implies C=2B$

$\implies\dfrac A1=\dfrac B2=\dfrac C4=\dfrac{A+B+C}{1+2+4}=\dfrac\pi7 $

Now use this or this