Show that \((a + s)(a + s)\) \(=\) \(a^2+ 2as + s^2\).
Proof:
\((a + s) (a + s)\) \(=\) \((a+s) a + (a + s)s\)
[By ]
\(=\) \((a × a) + (s × a) + (a \times s) + (s × s)\)
\(=\) \(a^{1+1} + sa + as + s^{1+1}\) [By ]
\(=\) \(a^{2} + as + as + a^{2}\) [Since ]
\(=\) \(a^{2} + 2as + s^{2}\) []
Therefore, \((a + s) (a + s)\) \(=\) \(a^{2} + 2as + s^{2}\).
Hence, proved.