Given: x=4aba+b
⇒x2a=2ba+b
by componendo and dividendo
⇒x+2ax−2a=2b+a+b2b−(a+b)
⇒x+2ax−2a=3b+a(b−a)....(i)
again x=4aba+b
⇒x2b=2ba+b
by componendo and dividendo
x+2bx−2b=2a+a+b2a−(a+b)
⇒x+2bx−2b=3a+b(a−b)
⇒x+2bx−2b=−(3a+b)b−a....(ii)
On adding (i) and (ii)
x+2ax−2a+x+2bx−2b=3b+ab−a−(3a+b)(b−a)
⇒x+2ax−2a+x+2bx−2b=3b+a−3a−b(b−a)
⇒x+2ax−2a+x+2bx−2b=2b−2ab−a
⇒x+2ax−2a+x+2bx−2b=2(b−a)(b−a)
⇒x+2ax−2a+x+2bx−2b=2