Label $\widehat A \equiv \measuredangle CAB, \widehat B \equiv \measuredangle ABC$. Let $D, E, F \in CA$ be points such that $\measuredangle ABD = \measuredangle DBE = \measuredangle EBF = \measuredangle FBC = \widehat A$ $\Longrightarrow$ $\triangle ABD$ is D-isosceles, $BE$ bisects $\measuredangle ABC$ and $BF$ bisects $\measuredangle EBC$. Label $l_b = [BE]$. $\triangle DBC \sim \triangle BEC$ $\Longrightarrow$ $\frac{c}{2a \cos \widehat A} = \frac{[BD]}{[BC]} =$ $\frac{[BE]}{[CE]} = \frac{l_b(c+a)}{ab}$ $\Longrightarrow$ $l_b = \frac{bc}{2(c+a) \cos \widehat A}$. $\triangle ABC \sim \triangle BFC$ $\Longrightarrow$ $\frac{b}{a} = \frac{[CA]}{[BC]} = \frac{[BC]}{[CF]} = a \cdot \frac{l_b + a}{a} \cdot \frac{c+a}{ab}$ $\Longrightarrow$ $l_b = \frac{b^2 - a(c+a)}{c+a}$. Comparing the 2 formulas for $l_b$ $\Longrightarrow$ $b^2 - a^2 - ca = \frac{bc}{2 \cos \widehat A}$. By sine theorem for $\triangle ABC$ $\Longrightarrow$ $\frac{b}{a} = \frac{[CA]}{[BC]} = \frac{\sin \widehat B}{\sin \widehat A} = \frac{\sin 4\widehat A}{\sin \widehat A} = 4 \cos \widehat A \cos 2\widehat A$ $\Longrightarrow$ $b^2 - a^2 - ca = \frac{bc}{2 \cos \widehat A} = 2 ca \cos 2\widehat A = 4 ca \cos ^2 \widehat A - 2 ca$ $\Longrightarrow$ $b^2 - a^2 + ca = 4 ca \cos ^2 \widehat A$. Combined $\Longrightarrow$ $(b^2 - a^2 - ca) \cdot \left[(b^2 - a^2)^2 - c^2a^2\right] = (b^2 - a^2 - ca)^2 \cdot (b^2 - a^2 + ca) =$ $\left(\frac{bc}{2 \cos \widehat A}\right)^2 \cdot 4 ca \cos ^2 \widehat A = ab^2c^3$.