Question-66599

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} \Rightarrow a^{2} - 2 a b + b^{2} = a^{2} - b^{2} \Rightarrow

2 b^{2} - 2 a b = 0 \Rightarrow 2 b (b - a) = 0 \Rightarrow {\begin{cases} b = 0 \\ a = b \end{cases}

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

{(a + b)}^{3} = a^{3} + 3 a b + b^{3} \Rightarrow

a^{3} + 3 {a b}^{2} + 3 a^{2} b + b^{3} = a^{3} + 3 a b + b^{3} \Rightarrow

3 a b (b + a - 1) = 0 {\begin{cases} a b = 0 \\ a + b - 1 = 0 \Rightarrow a + b = 1 \end{cases}