Question Number 66599 by aliesam last updated on 17/Aug/19
Commented by kaivan.ahmadi last updated on 17/Aug/19
$$\left({a}−{b}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} −{b}^{\mathrm{2}} \Rightarrow{a}^{\mathrm{2}} −\mathrm{2}{ab}+{b}^{\mathrm{2}} ={a}^{\mathrm{2}} −{b}^{\mathrm{2}} \Rightarrow \\ $$$$\mathrm{2}{b}^{\mathrm{2}} −\mathrm{2}{ab}=\mathrm{0}\Rightarrow\mathrm{2}{b}\left({b}−{a}\right)=\mathrm{0}\Rightarrow\begin{cases}{{b}=\mathrm{0}}\\{{a}={b}}\end{cases} \\ $$
Commented by kaivan.ahmadi last updated on 17/Aug/19
$$\left({a}+{b}\right)^{\mathrm{3}} ={a}^{\mathrm{3}} +\mathrm{3}{ab}+{b}^{\mathrm{3}} \Rightarrow \\ $$$${a}^{\mathrm{3}} +\mathrm{3}{ab}^{\mathrm{2}} +\mathrm{3}{a}^{\mathrm{2}} {b}+{b}^{\mathrm{3}} ={a}^{\mathrm{3}} +\mathrm{3}{ab}+{b}^{\mathrm{3}} \Rightarrow \\ $$$$\mathrm{3}{ab}\left({b}+{a}−\mathrm{1}\right)=\mathrm{0\begin{cases}{{ab}=\mathrm{0}}\\{{a}+{b}−\mathrm{1}=\mathrm{0}\Rightarrow{a}+{b}=\mathrm{1}}\end{cases}} \\ $$