P=$\frac{a³}{(a-b).(a-c)}$ + $\frac{b³}{(b-a).(b-c)}$ + $\frac{c³}{(c-a).(c-b)}$

= $\frac{a³.(b-c)}{(a-b).(a-c).(b-c)}$-$\frac{b³.(a-c)}{(a-b).(a-c).(b-c)}$ + $\frac{c³.(a-b)}{(a-b).(a-c).(b-c)}$ = $\frac{a³.(b-c)-b³.(a-c)+c³.(a-b)}{(a-b).(a-c).(b-c)}$ 

Xét ở tử số, ta có: a³.(b-c)-b³.(a-c)+c³.(a-b)

= a³.( b-c)-ab³+b³c+ac³-bc³

= a³.( b-c)-a.( b-c).( b²+bc+c²)+bc.( b-c).( b+c)

= ( b-c).( a³-ab²-abc-ac²+b²c+bc²)

= ( b-c).[ a.( a-b).( a+b)-bc.( a-b)-c².( a-b)

= ( a-b).( b-c).( a²+ab-bc-c²)

= ( a-b).( b-c).[(a-c).(a+c)+b(a-c)]

= ( a-b).( b-c).( a-c).( a+b+c)

⇒ P= $\frac{( a-b).( b-c).( a-c).( a+b+c)}{(a-b).(a-c).(b-c)}$= a+b+c

Vì a, b, c đều là số nguyên nên P cũng là số nguyên