Докажите тождество (с+d)^2 (c-d)+6cd (c-d)+10cd (d-c)=(c-d)^3 ! ^-^

(с+d)² ·(c-d)+6cd (c-d)+10cd (d-c)=(c-d)³ 
(с+d)² ·(c-d)+6cd (d-с)-10cd (c-d)=(c-d)³ 
(c-d)·((c+d)²-4cd) = (c-d)³
(c-d)·(c²+2cd+d²-4cd) = (c-d)³
(c-d)·(c²-2cd+d²) = (c-d)³
(c-d)·(c-d)² = (c-d)³
(c-d)³ = (c-d)³