Question
If $\text{x}=\text{a}^{\text{m}+\text{n}},\ \text{y}=\text{a}^{\text{n}+\text{l}}$ and $\text{z}=\text{a}^{\text{l}+\text{m}},$ prove that $\text{x}^\text{m}\text{y}^\text{n}\text{z}^\text{l}=\text{x}^\text{n}\text{y}^\text{l}\text{z}^\text{m}.$

Answer

$\text{x}=\text{a}^{\text{m}+\text{n}},\ \text{y}=\text{a}^{\text{n}+\text{l}}$ and $\text{z}=\text{a}^{\text{l}+\text{m}},$
$\text{LHS}=\text{x}^\text{m}\text{y}^\text{n}\text{z}^\text{l}$
$=\big(\text{a}^{\text{m}+\text{n}}\big)^\text{m}\times\big(\text{a}^{\text{n}+\text{l}}\big)^{\text{n}}\times\big(\text{a}^{\text{l}+\text{m}}\big)^\text{l}$
$=\text{a}^{\text{m}^{2}+\text{mn}}\times\text{a}^{\text{n}^2+\text{nl}}\times\text{a}^{\text{l}^2+\text{lm}}$
$=\text{a}^{\text{m}^2+\text{mn}+\text{n}^2+\text{nl}+\text{l}^2+\text{lm}}$
$=\text{a}^{\text{mn}+\text{n}^2}\times\text{a}^{\text{nl}+\text{l}^2}\times\text{a}^{\text{lm}+\text{m}^2}$
$=\text{a}^{\text{n}(\text{m}+\text{n})}\times\text{a}^{\text{l}(\text{n}+\text{l})}\times\text{a}^{\text{m}(\text{l}+\text{m})}$
$=\big(\text{a}^{(\text{m}+\text{n})}\big)^\text{n}\times\big(\text{a}^{(\text{n}+\text{l})}\big)^\text{l}\times\big(\text{a}^{(\text{l}+\text{m})}\big)^\text{m}$
$=\text{x}^\text{n}\text{y}^\text{l}\text{z}^\text{m}$

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