Let $ABC$ be a triangle with angles $\angle A = 80^\circ, \angle B = 70^\circ, \angle C = 30^\circ$. Let $P$ be a point on the bisector of $\angle BAC$ satisfying $\angle BPC =130^\circ$. Let $PX, PY, PZ$ be the perpendiculars drawn from $P$ to the sides $BC, AC, AB$, respectively. Prove that the following equation with segment lengths is satisfied $$AY^3+BZ^3+CX^3=AZ^3+BX^3+CY^3.$$