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-1} c^{l-m}=1 . $

Given:

\( a=x^{m+n} y^{l}, b=x^{n+l} y^{m} \) and \( c=x^{l+m} y^{n} \)

To do: 

We have to prove that \( a^{m-n} b^{n-1} c^{l-m}=1 . \)

Solution:

We know that,

$(a^{m})^{n}=a^{m n}$

$a^{m} \times a^{n}=a^{m+n}$

$a^{m} \div a^{n}=a^{m-n}$

$a^{0}=1$

Therefore,

LHS $=a^{m-n} b^{n-l} c^{l-m}$

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

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

$=x^{m^2-n^2} y^{lm-ln} x^{n^2-l^2} y^{mn-ml} x^{l^2-m^2} y^{nl-nm}$

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

$=x^0 \times y^0$

$=1\times1$

$=1$

$=$ RHS

Hence proved.