We have,
(i)16a2–254a2
=(4a)2−(52a)2=(4a+52a)(4a−52a)
(ii)16a2b–b16a2=b(16a2−116a2)
=b{(4a)2−(14a)2}
=b(4a+14a)(4a−14a)
(iii)100(x+y)2–81(a+b)2={10(x+y)}2–{9(a+b)}2
={10(x+y)+9(a+b)}{10(x+y)–9(a+b)}
=(10x+10y+9a+9b)(10x+10y–9a–9b)
(iv)(x–1)2–(x–2)2={(x–1)+(x–2)}{(x–1)–(x–2)}
=(2x–3)(x–1–x+2)
=(2x–3)×1
=2x–3