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Question Number 66599 by aliesam last updated on 17/Aug/19

Commented by kaivan.ahmadi last updated on 17/Aug/19

(a−b)^2 =a^2 −b^2 ⇒a^2 −2ab+b^2 =a^2 −b^2 ⇒ 2b^2 −2ab=0⇒2b(b−a)=0⇒ { ((b=0)),((a=b)) :}

$$\left({a}−{b}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} −{b}^{\mathrm{2}} \Rightarrow{a}^{\mathrm{2}} −\mathrm{2}{ab}+{b}^{\mathrm{2}} ={a}^{\mathrm{2}} −{b}^{\mathrm{2}} \Rightarrow \\ $$$$\mathrm{2}{b}^{\mathrm{2}} −\mathrm{2}{ab}=\mathrm{0}\Rightarrow\mathrm{2}{b}\left({b}−{a}\right)=\mathrm{0}\Rightarrow\begin{cases}{{b}=\mathrm{0}}\\{{a}={b}}\end{cases} \\ $$

Commented by kaivan.ahmadi last updated on 17/Aug/19

(a+b)^3 =a^3 +3ab+b^3 ⇒ a^3 +3ab^2 +3a^2 b+b^3 =a^3 +3ab+b^3 ⇒ 3ab(b+a−1)=0 { ((ab=0)),((a+b−1=0⇒a+b=1)) :}

$$\left({a}+{b}\right)^{\mathrm{3}} ={a}^{\mathrm{3}} +\mathrm{3}{ab}+{b}^{\mathrm{3}} \Rightarrow \\ $$$${a}^{\mathrm{3}} +\mathrm{3}{ab}^{\mathrm{2}} +\mathrm{3}{a}^{\mathrm{2}} {b}+{b}^{\mathrm{3}} ={a}^{\mathrm{3}} +\mathrm{3}{ab}+{b}^{\mathrm{3}} \Rightarrow \\ $$$$\mathrm{3}{ab}\left({b}+{a}−\mathrm{1}\right)=\mathrm{0\begin{cases}{{ab}=\mathrm{0}}\\{{a}+{b}−\mathrm{1}=\mathrm{0}\Rightarrow{a}+{b}=\mathrm{1}}\end{cases}} \\ $$

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