Lời giải:
Ta có:
\((a+b)^5=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5\)
\(=(a^5+b^5)+5ab(a^3+2a^2b+2ab^2+b^3)\)
\(=a^5+b^5+5ab[(a+b)(a^2-ab+b^2)+2ab(a+b)]\)
\(=a^5+b^5+5ab(a+b)(a^2+ab+b^2)\)
Áp dụng vào bài toán:
\((a+b+c)^5-a^5-b^5-c^5\)
\(=a^5+(b+c)^5+5a(b+c)(a+b+c)[a^2+a(b+c)+(b+c)^2]-a^5-b^5-c^5\)
\(=(b+c)^5+5a(b+c)(a+b+c)(a^2+b^2+c^2+ab+ac+2bc)-b^5-c^5\)
\(=b^5+c^5+5bc(b+c)(b^2+bc+c^2)+5a(b+c)(a+b+c)(a^2+b^2+c^2+ab+ac+2bc)-b^5-c^5\)
\(=5bc(b+c)(b^2+bc+c^2)+5a(b+c)(a+b+c)(b^2+bc+c^2)+5a(b+c)(a+b+c)(a^2+ab+ac+bc)\)
\(=5(b+c)(b^2+bc+c^2)(bc+a^2+ab+ac)+5a(b+c)(a+b+c)(a^2+ab+ac+bc)\)
\(=5(b+c)(a^2+ab+bc+ac)(b^2+bc+c^2+a^2+ab+ac)\)
\(=5(b+c)(a+b)(a+c)(b^2+bc+c^2+a^2+ab+ac)\)
