I tried to use the reciprocity law to prove it, but neither $a$ nor $bc$ is odd prime.
Then, I tried by definition,
${\large{a\choose bc}}=a^{\large{\frac{bc-1}{2}}} \pmod{bc}$.
${\large{a\choose b}{a\choose c}}=a^{\large{\frac{b-1}{2}}}a^{\large{\frac{c-1}{2}}}=a^{\large{\frac{bc-b-c+1}{4}}}\pmod{bc}$.
Now, I need to prove $$a^{\large{\frac{bc-1}{2}}}=a^{\large{\frac{bc-b-c+1}{4}}}\pmod{bc}$$ and I am clueless...
The question suggests the problem may be some confusion about Jacobi vs Legendre symbols. Legendre symbols $(\frac{a}{b})$ are defined for any integer $a$ and odd primes $b$. Then for odd $n$, the Jacobi symbols are defined as a product of Legendre symbols
$$(\frac{a}{n}) = (\frac{a}{p_1})^{\nu_1}\cdot\ldots\cdots(\frac{a}{p_k})^{\nu_k},$$
where $n=\prod_{i=1}^k p_i^{\nu_i}$ for odd primes $p_i$. In order to say this, we of course need to have that $n$ is odd.
Now for the question, we need to write $b$ and $c$ as product of primes to get something like $$(\frac{a}{bc})=\prod_{p_i|bc}(\frac{a}{p_i})^{\nu_i}=\prod_{p_i|b}(\frac{a}{p_i})^{\mu_i}\cdot\prod_{p_i|c}(\frac{a}{p_i})^{\kappa_i} = (\frac{a}{b})\cdot(\frac{a}{c}).$$
A bit of detail is still needed for the $\mu$'s and $\kappa$'s, but that should not be too hard to figure out. Is this more or less what you were looking for?
(Also, if anyone knows how to format this somewhat neater, please do so! Somehow \large doesn't quite do what I want it to do.)