If `x=a^(m+n),\ y=a^(n+l)` and `z=a^(l+m)` , prove that `x^m y^n z^l=x^n y^l z^m`

Explanation

`x^my^nz^l`
`=(a^(m+n))^mxx(a^(n+l))^nxx(a^(l+m))^l`
`=a^(m^2+mn)xxa^(n^2+ln)xxa^(l^2+ml)`
`=a^(m^2+mn+n^2+ln+l^2+ml)` ...........(1)
`x^ny^lz^m`
`=(a^(m+n))^nxx(a^(n+l))^lxx(a^(l+m))^m`
`=a^(mn+n^2)a^(nl+l^2)a^(lm+m^2)`
`=a^(mn^+n^2+nl+l^2+lm+m^2)` ...........(2)
On comparing (1) and (2), its clear that,
`x^my^nz^l=x^ny^lz^m`
Hence Proved.