Question

If `a=x^(m+n)y^l, b=x^(n+l)y^m` and `c=x^(l+m)y^n,` Prove that `a^(m-n)b^(n-l)c^(l-m)=1`

Answer 1

Given `a=x^(m+n)y^l, b=x^(n+l)y^m` and `c=x^(l+m)y^n`

Putting the values of a, b and c in `a^(m-n)b^(n-l)c^(l-m),` we get

`a^(m-n)b^(n-l)c^(l-m)`

`=(x^(m+n)y^l)^(m-n)(x^(n+l)y^m)^(n-l)(x^(l+m)y^n)^(l-m)`

`=[x^((m+n)(m-n))y(l(m-n))][x^((n+l)(n-l))y^(m(n-l))][x^((l+m)(l_m))y^(n(l-m))]`

`=x^((m^2-n^2))x^((n^2-l^2))x^((l^2-m^2))y^(lm-ln)y^(mn-ml)y^(nl-nm)`

`=x^0y^0`

= 1