2003年第14届希望杯初二竞赛第1试第2题
若$$x+y=-1$$，则$${{x}^{4}}+5{{x}^{3}}y+{{x}^{2}}y+8{{x}^{2}}{{y}^{2}}+x{{y}^{2}}+5x{{y}^{3}}+{{y}^{4}}$$的值等于（~ ）．
课内体系->知识点->式->整式的乘除->整式乘除化简求值->整式乘除化简求值-整体代入化简求值 课内体系->能力->运算能力
$${{x}^{4}}+5{{x}^{3}}y+{{x}^{2}}y+8{{x}^{2}}{{y}^{2}}+x{{y}^{2}}+5x{{y}^{3}}+{{y}^{4}}$$  $$=({{x}^{4}}+2{{x}^{2}}{{y}^{2}}+{{y}^{4}})+(5{{x}^{3}}y+5x{{y}^{3}})+6{{x}^{2}}{{y}^{2}}+{{x}^{2}}y+x{{y}^{2}}$$  $$={{({{x}^{2}}+{{y}^{2}})}^{2}}+5xy({{x}^{2}}+{{y}^{2}})+6{{x}^{2}}{{y}^{2}}+{{x}^{2}}y+x{{y}^{2}}$$  $$=({{x}^{2}}+{{y}^{2}}+2xy)({{x}^{2}}+{{y}^{2}}+3xy)+xy(x+y)$$  $$={{(x+y)}^{2}}({{x}^{2}}+{{y}^{2}}+3xy)+xy(x+y)$$  $$={{x}^{2}}+{{y}^{2}}+3xy-xy$$  $$={{x}^{2}}+{{y}^{2}}+2xy$$  $$={{(x+y)}^{2}}=1$$． 