恒等式の証明（展開して比べる）（B）

(1) \[ (\text{左辺})=x^2y^2-x^2-y^2+1 \] \[ (\text{右辺})=(x^2y^2+2xy+1)-(x^2+2xy+y^2)=x^2y^2-x^2-y^2+1 \] よって $(\text{左辺})=(\text{右辺})$ (2) \[ (\text{左辺})=a^4+b^4+(a^4+4a^3b+6a^2b^2+4ab^3+b^4)=2a^4+4a^3b+6a^2b^2+4ab^3+2b^4 \] \[ (a^2+ab+b^2)^2=a^4+a^2b^2+b^4+2a^3b+2ab^3+2a^2b^2=a^4+2a^3b+3a^2b^2+2ab^3+b^4 \] であるから \[ (\text{右辺})=2a^4+4a^3b+6a^2b^2+4ab^3+2b^4=(\text{左辺}) \] \hfill 〔証明終〕