2000年第11届希望杯初二竞赛第2试第15题
$${{a}^{2}}-{{b}^{2}}=1+\sqrt{2}$$，$${{b}^{2}}-{{c}^{2}}=1-\sqrt{2}$$, find $${{a}^{4}}+{{b}^{4}}+{{c}^{4}}-{{a}^{2}}{{b}^{2}}-{{b}^{2}}{{c}^{2}}-{{c}^{2}}{{a}^{2}}$$.  设$${{a}^{2}}-{{b}^{2}}=1+\sqrt{2}$$，$${{b}^{2}}-{{c}^{2}}=1-\sqrt{2}$$，则$${{a}^{4}}+{{b}^{4}}+{{c}^{4}}-{{a}^{2}}{{b}^{2}}-{{b}^{2}}{{c}^{2}}-{{c}^{2}}{{a}^{2}}=$$~\uline{~~~~~~~~~~}~．
竞赛->知识点->数与式->整式->乘法公式 课内体系->知识点->式->整式的加减->整式有关的概念->整式
因为$${{a}^{2}}-{{b}^{2}}=1+\sqrt{2}$$，  $${{b}^{2}}-{{c}^{2}}=1-\sqrt{2}$$，  所以$${{a}^{2}}-{{c}^{2}}=2$$，  所以$${{a}^{4}}+{{b}^{4}}+{{c}^{4}}-{{a}^{2}}{{b}^{2}}-{{b}^{2}}{{c}^{2}}-{{c}^{2}}{{a}^{2}}$$  $$=\frac{1}{2}[({{a}^{4}}-2{{a}^{2}}{{b}^{2}}+{{b}^{4}})+({{b}^{4}}-2{{b}^{2}}{{c}^{2}}+{{c}^{4}})+({{c}^{4}}-2{{a}^{2}}{{c}^{2}}+{{a}^{4}})]$$  $$=\frac{1}{2}[{{({{a}^{2}}-{{b}^{2}})}^{2}}+{{({{b}^{2}}-{{c}^{2}})}^{2}}+{{({{c}^{2}}-{{a}^{2}})}^{2}}]$$  $$=\frac{1}{2}[{{(1+\sqrt{2})}^{2}}+{{(1-\sqrt{2})}^{2}}+{{2}^{2}}]$$  $$=5$$． 