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Hint: Estimate L.H.S and R.H.S separately, and use this property $\left( {{p^b} \cdot {p^c} \cdot {p^d} = {p^{b + c + d}}} \right)$

We have to prove ${x^m}{y^n}{z^l} = {x^n}{y^l}{z^m}$

Let’s take L.H.S which is ${x^m}{y^n}{z^l}..........\left( 1 \right)$

It is given that $x = {a^{m + n}},y = {a^{n + l}},z = {a^{l + m}}$

So, substitute these values in expression (1)

$

\Rightarrow {x^m}{y^n}{z^l} = {a^{\left( {m + n} \right)}}^m{a^{\left( {n + l} \right)n}}{a^{\left( {l + m} \right)l}} \\

{\text{ }} = {a^{{m^2} + nm}}{a^{{n^2} + nl}}{a^{{l^2} + ml}} \\

$

In the R.H.S. of the above equation, the base is the same that is $a$ and it is a product, hence the powers will add up.

$

\Rightarrow {x^m}{y^n}{z^l} = {a^{{m^2} + nm}}{a^{{n^2} + nl}}{a^{{l^2} + ml}} \\

{\text{ }} = {a^{{m^2} + {n^2} + {l^2} + lm + mn + nl}}.................\left( 2 \right) \\

{\text{ }} \\

$

Now let’s take R.H.S of the equation we have to prove, which is ${x^n}{y^l}{z^m}...........\left( 3 \right)$

It is given that $x = {a^{m + n}},y = {a^{n + l}},z = {a^{l + m}}$

So, substitute these values in expression (3)

$

\Rightarrow {x^n}{y^l}{z^m} = {a^{\left( {m + n} \right)}}^n{a^{\left( {n + l} \right)l}}{a^{\left( {l + m} \right)m}} \\

{\text{ }} = {a^{{n^2} + nm}}{a^{{l^2} + nl}}{a^{{m^2} + ml}} \\

$

In the R.H.S. of the above equation, the base is the same that is $a$ and it is a product, hence the powers will add up.

$

\Rightarrow {x^n}{y^l}{z^m} = {a^{{n^2} + nm}}{a^{{l^2} + nl}}{a^{{m^2} + ml}} \\

{\text{ }} = {a^{{m^2} + {n^2} + {l^2} + lm + mn + nl}}.................\left( 4 \right) \\

{\text{ }} \\

$

Now from equation (2) and (4) it is clear that for the main equation the L.H.S = R.H.S $ = {a^{{m^2} + {n^2} + {l^2} + lm + mn + nl}}$

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

Hence proved

Note: - In these types of questions the key concept is that if the base is the same and all the terms are a product, then the powers of the bases will add up, then we can simplify L.H.S and R.H.S separately using this property to obtain the required answer.

We have to prove ${x^m}{y^n}{z^l} = {x^n}{y^l}{z^m}$

Let’s take L.H.S which is ${x^m}{y^n}{z^l}..........\left( 1 \right)$

It is given that $x = {a^{m + n}},y = {a^{n + l}},z = {a^{l + m}}$

So, substitute these values in expression (1)

$

\Rightarrow {x^m}{y^n}{z^l} = {a^{\left( {m + n} \right)}}^m{a^{\left( {n + l} \right)n}}{a^{\left( {l + m} \right)l}} \\

{\text{ }} = {a^{{m^2} + nm}}{a^{{n^2} + nl}}{a^{{l^2} + ml}} \\

$

In the R.H.S. of the above equation, the base is the same that is $a$ and it is a product, hence the powers will add up.

$

\Rightarrow {x^m}{y^n}{z^l} = {a^{{m^2} + nm}}{a^{{n^2} + nl}}{a^{{l^2} + ml}} \\

{\text{ }} = {a^{{m^2} + {n^2} + {l^2} + lm + mn + nl}}.................\left( 2 \right) \\

{\text{ }} \\

$

Now let’s take R.H.S of the equation we have to prove, which is ${x^n}{y^l}{z^m}...........\left( 3 \right)$

It is given that $x = {a^{m + n}},y = {a^{n + l}},z = {a^{l + m}}$

So, substitute these values in expression (3)

$

\Rightarrow {x^n}{y^l}{z^m} = {a^{\left( {m + n} \right)}}^n{a^{\left( {n + l} \right)l}}{a^{\left( {l + m} \right)m}} \\

{\text{ }} = {a^{{n^2} + nm}}{a^{{l^2} + nl}}{a^{{m^2} + ml}} \\

$

In the R.H.S. of the above equation, the base is the same that is $a$ and it is a product, hence the powers will add up.

$

\Rightarrow {x^n}{y^l}{z^m} = {a^{{n^2} + nm}}{a^{{l^2} + nl}}{a^{{m^2} + ml}} \\

{\text{ }} = {a^{{m^2} + {n^2} + {l^2} + lm + mn + nl}}.................\left( 4 \right) \\

{\text{ }} \\

$

Now from equation (2) and (4) it is clear that for the main equation the L.H.S = R.H.S $ = {a^{{m^2} + {n^2} + {l^2} + lm + mn + nl}}$

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

Hence proved

Note: - In these types of questions the key concept is that if the base is the same and all the terms are a product, then the powers of the bases will add up, then we can simplify L.H.S and R.H.S separately using this property to obtain the required answer.