Home > Practice > Arithmetic Aptitude > Surds and Indices > Miscellaneous
1. (xbxc)(b+ca).   (xcxa)(c+ab).   (xaxb)(a+bc)   = ?

Solution:
Given Exp.=x(bc)(b+ca).x(ca)(c+ab).x(ab)(a+bc)
  =x(bc)(b+c)a(bc).    x(ca)(c+a)b(ca).   x(ab)(a+b)c(ab)
=x(b2c2+c2a2+a2b2).xa(bc)b(ca)c(ab)=x0.x(ab+acbc+baca+cb)=(x0×x0)=(1×1)=1

You must login to add comments. Login now.