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

$a=x^{m+n}.y^l$

$a^{m-n}=(x^{m+n}.y^l{)}^{m-n}$

$a^{m-n}=x^{m^2-n^2}.y^{ml-nl}$

$b=x^{n+l}.y^m$

$b^{n-l}=(x^{n+l}{y}^m{)}^{n-l}$

$b^{n-l}=x^{n^2-l^2}.y^{mn-ml}$

$c=x^{l+m}.y^n$

$c^{l-m}=(x^{l+m}.y^n{)}^{l-m}$

$c^{l-m}=x^{l^2-m^2}.y^{nl-mn}$

$a^{m-n}.b^{n-l}.c^{l-m}=x^{m^2-n^2+n^2-l^2+l^2-m^2}.y^{ml-nl+mn-ml+nl-mn}$

$=x^0.y^0$

$=1$