Points $P$ and $Q$ inside a triangle $ABC$ with sides $a,b,c$ and the corresponding angle $\alpha,\beta,\gamma$ satisfy $\angle BPC=\angle CPA=\angle APB=120^\circ$ and $\angle BQC=60^\circ+\alpha$, $\angle CQA=60^\circ+\beta$, $\angle AQB=60^\circ+\gamma$. Prove the equality $$(AP+BP+CP)^3\cdot AQ\cdot BQ\cdot CQ=(abc)^2.$$