Let $[{\bf a} \times {\bf b} \space \; {\bf b} \times {\bf c}\space\; {\bf c} \times {\bf a} ]=k[{\bf a}\space{\bf b}\space{\bf c}]^2$. Find $k$. Here, $[{\bf u}\ {\bf v} \ {\bf w}]={\bf u}\cdot({\bf v}\times {\bf w})$.
My attempt:
Writing scalar triple product as : $({\bf a} \times {\bf b}) \cdot \big(({\bf b} \times {\bf c}) \times ({\bf c}\times {\bf a})\big)$. Not able to proceed next.
As hinted in a comment, we use the identity ${\bf u}\times({\bf v}\times {\bf w})=({\bf u}\cdot {\bf w}){\bf v}-({\bf u}\cdot {\bf v}){\bf w}$ (I proved it here). Let $$\Delta={\bf a}\cdot({\bf b}\times {\bf c})={\bf b}\cdot({\bf c}\times {\bf a})={\bf c}\cdot({\bf a}\times {\bf b}).$$ First we compute $$({\bf b}\times {\bf c})\times ({\bf c}\times {\bf a})=\big(({\bf b}\times {\bf c})\cdot {\bf a}\big){\bf c}-\big(({\bf b}\times {\bf c})\cdot {\bf c}){\bf a}=\Delta {\bf c}-0{\bf a}=\Delta {\bf c}.$$ Therefore $$({\bf a}\times {\bf b})\cdot\big(({\bf b}\times {\bf c})\times ({\bf c}\times {\bf a})\big)=({\bf a}\times {\bf b})\cdot \Delta {\bf c}=\Delta\big(({\bf a}\times {\bf b})\cdot {\bf c}\big)=\Delta^2.$$ Hence $k=1$.