show that if $\displaystyle {a}^{2}+{b}^{2}={c}^{2}$, then

$\displaystyle {log}_{b+c}a + {log}_{c-b}a= 2\left( {log}_{b+c}a\right)\left( {log}_{c-b}a \right) $

I tried to solve this with change of base but didn't seem to help. so not sure even what the first step is on this one.