Đặt [tex]p=a+b+c,q=ab+bc+ca,r=abc[/tex]
[tex]2q=(a+b+c)^2-(a^2+b^2+c^2)=-20\Rightarrow q=-10[/tex]
Lại có: [tex](a+b+c)^5-a^5-b^5-c^5=5(a+b)(b+c)(c+a)(a^2+b^2+c^2+ab+bc+ca)=5(pq-r)\frac{1}{2}(p^2+q)\Rightarrow a^5+b^5+c^5=p^5-\frac{5}{2}(pq-r)(p^2+q)[/tex]