Question

x raise to the power m+n* x raise to the power n+l * x raise to the power l+m / (x raise to the power m * x raise to the power n * x raise to the power l)² = 1

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Answer 1

$\large\underline{\sf{Given \:Question - }}$

Prove that,

$\rm :\longmapsto\:\dfrac{ {x}^{l + m} \times {x}^{m + n} \times {x}^{n + l} }{ {( {x}^{l} \times {x}^{m} \times {x}^{n} )}^{2} } = 1$

$\large\underline{\sf{Solution-}}$

Consider LHS

$\rm :\longmapsto\:\dfrac{ {x}^{l + m} \times {x}^{m + n} \times {x}^{n + l} }{ {( {x}^{l} \times {x}^{m} \times {x}^{n} )}^{2} }$

Let we first evaluate numerator,

$\red{\rm :\longmapsto\:{x}^{l + m} \times {x}^{m + n} \times {x}^{n + l}}$

We know,

$\purple{\boxed{ \bf{ {x}^{m} \times {x}^{n} = {x}^{m + n}}}}$

So, using this identity, we get

$\rm \:  =  \:  \: {x}^{l + m + m + n + n + l}$

$\rm \:  =  \:  \: {x}^{2l + 2m + 2n}$

$\bf\implies \:{x}^{l + m} \times {x}^{m + n} \times {x}^{n + l} \: = \: {x}^{2l + 2m + 2n} - - (1)$

Now, let we evaluate the denominator,

$\red{\rm :\longmapsto\:{( {x}^{l} \times {x}^{m} \times {x}^{n} )}^{2}}$

We know,

$\purple{\boxed{ \bf{ {x}^{m} \times {x}^{n} = {x}^{m + n}}}}$

So using this identity, we get

$\rm \:  =  \:  \: {( {x}^{l + m + n} )}^{2}$

We know,

$\purple{\boxed{ \bf{ {( {x}^{m} )}^{n} = {x}^{mn}}}}$

So, using this identity, we get

$\rm \:  =  \:  \: {x}^{2l + 2m + 2n}$

$\bf\implies \:{( {x}^{l} \times {x}^{m} \times {x}^{n} )}^{2} \: = \: {x}^{2l + 2m + 2n} - - - (2)$

Now, Consider,

$\bf :\longmapsto\:\dfrac{ {x}^{l + m} \times {x}^{m + n} \times {x}^{n + l} }{ {( {x}^{l} \times {x}^{m} \times {x}^{n} )}^{2} }$

On substituting the values from equation (1) and equation (2), we get

$\rm \:  =  \:  \: \dfrac{ {x}^{2l + 2m + 2n} }{{x}^{2l + 2m + 2n}}$

$\rm \:  =  \:  \: 1$

Hence,

$\bf :\longmapsto\:\dfrac{ {x}^{l + m} \times {x}^{m + n} \times {x}^{n + l} }{ {( {x}^{l} \times {x}^{m} \times {x}^{n} )}^{2} } = 1$

Additional Information :-

$\boxed{ \rm{ {x}^{m} \div {x}^{n} = {x}^{m - n}}}$

$\boxed{ \rm{ {x}^{0} = 1}}$

$\boxed{ \rm{ {x}^{ - y} = \frac{1}{ {x}^{y} }}}$