If \( x=a^{m+n}, y=a^{n+1} \) and \( z=a^{l+m} \), prove that \( x^{m} y^{n} z^{l}=x^{n} y^{l} z^{m} \)

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 \( x^{m} y^{n} z^{l}=x^{n} y^{l} z^{m} \)

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 $=x^{m} y^{n} z^{l}$

$=a^{(m+n) m} \times a^{(n+l) n} \times a^{(l+m)l}$

$=a^{m^{2}+m n} \times a^{n^{2}+n l} \times a^{l^{2}+m l}$

$=a^{m^{2}+n^{2}+l^{2}+m n+n l+l m}$

RHS $=x^{n} y^{l} z^{m}$

$=(a^{m+n})^{n} \times (a^{n+l})^{l} \times (a^{l+m})^{m}$

$=a^{m n+n^{2}} \times a^{n l+l^{2}} \times a^{l m+m^{2}}$

$=a^{m n+n^{2}+n l+l^{2}+l m+m^{2}}$

$=a^{l^{2}+m^{2}+n^{2}+l m+m n+n l}$

LHS $=$ RHS

Hence proved. 

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Updated on: 10-Oct-2022

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