Question

how I can solve this question

Answer 1

hope this helps any doubts please comment

Answer 2

Let us first consider $a^{m-n}$

 $a^{m-n}$ = $( x^{m+n} y^{l} ) ^ {m-n}$
= $( x^{m+n})^{m-n} ( y^{l} ) ^ {m-n}$
$= ( x^{m^{2}-n^{2}}) ( y^{ml-nl} )$

Now, coming to $b^{n-l}$

$b^{n-l}$ = $( x^{n+l} y^{m} ) ^ {n-l}$
= $( x^{n+l})^{n-l} ( y^{m} ) ^ {n-l}$
$= ( x^{n^{2}-l^{2}}) ( y^{mn-ml} )$

At last, coming to $c^{l-m}$

$c^{l-m}$ = $( x^{l+m} y^{n} ) ^ {l-m}$
= $( x^{l+m})^{l-m} ( y^{n} ) ^ {l-m}$
$= ( x^{l^{2}-m^{2}}) ( y^{nl-mn} )$

Now, multiplying the three results, we get
$a^{m-n}$$b^{n-l}$ $c^{l-m}$

= $( x^{m^{2}-n^{2}}) ( y^{ml-nl} ) * ( x^{n^{2}-l^{2}}) ( y^{mn-ml} ) * ( x^{l^{2}-m^{2}}) ( y^{nl-mn} )$

= $( x^{m^{2}-n^{2}+n^{2}-l^{2}+l^{2}-m^{2}}) ( y^{ml-nl+mn-ml+nl-mn} )$

=$x^{0} y^{0}$
= 1
= RHS

Hence proved.