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