Step-by-step explanation:

m = ax, n = ay

and a2 = (my.nx)z = myz.nxz = (ax)yz.(ay)xz = axyz.axyz = a2xyz

a2 = a2xyz

comparing the powers

⇒ 2 = 2xyz

⇒ xyz = 1

Verified answer

[tex]\large\underline{\sf{Solution-}}[/tex]

Given that,

[tex]\sf \: m = {a}^{x} - - - (1) \\ \\ [/tex]

[tex]\sf \: n = {a}^{y} - - - (2) \\ \\ [/tex]

and

[tex]\sf \: {( {m}^{y} \: {n}^{x}) }^{z} = {a}^{2} \\ \\ [/tex]

can be rewritten as

[tex]\sf \: {m}^{yz} \: {n}^{xz} = {a}^{2} \\ \\ [/tex]

[tex]\boxed{ \sf{ \: \because \: {(xy)}^{n} \: = \: {x}^{n} \times {y}^{n} \: }} \\ \\ [/tex]

On substituting the value of m and n from equation (1) and (2), we get

[tex]\sf \: {( {a}^{x} )}^{yz} \: {( {a}^{y}) }^{xz} = {a}^{2} \\ \\ [/tex]

[tex]\sf \: {a}^{xyz} \: \times \: {a}^{xyz} = {a}^{2} \\ \\ [/tex]

[tex]\boxed{ \sf{ \: \because \: {( {x}^{m} )}^{n} = {x}^{mn} \: }} \\ \\ [/tex]

[tex]\sf \: {a}^{2xyz} \: = \: {a}^{2} \\ \\ [/tex]

[tex]\boxed{ \sf{ \: \because \: {x}^{m} \times {x}^{n} = {x}^{m + n} \: }} \\ \\ [/tex]

[tex]\sf \:  \implies \: 2xyz = 2 \\ \\ [/tex]

[tex]\sf \:  \implies \: xyz = 1 \\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional information

[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: Formulae}}}} \\ \\ \bigstar \: \bf{ {x}^{0} = 1}\\ \\ \bigstar \: \bf{ {x}^{m} \times {x}^{n} = {x}^{m + n} }\\ \\ \bigstar \: \bf{ {( {x}^{m})}^{n} = {x}^{mn} }\\ \\\bigstar \: \bf{ {x}^{m} \div {x}^{n} = {x}^{m - n} }\\ \\ \bigstar \: \bf{ {x}^{ - n} = \dfrac{1}{ {x}^{n} } }\\ \\\bigstar \: \bf{ {\bigg(\dfrac{a}{b} \bigg) }^{ - n} = {\bigg(\dfrac{b}{a} \bigg) }^{n} }\\ \\\bigstar \: \bf{ {x}^{m} = {x}^{n}\rm\implies \:m = n }\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]