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If the two powers have the bases same then there will be the multiplication of the powers, so we will add the exponents.

${x^a} \cdot {x^b} = {x^{a + b}}$

Here, $x$ will be the base

And $a,b$ will be the exponents.

So we have the power given as $\dfrac{{{x^{m + n}} \times {x^{n + 1}} \times {x^{1 + m}}}}{{{{\left( {{x^m} \times {x^n} \times {x^1}} \right)}^2}}} = 1$ .

Now on taking the numerator part of the LHS side of the equation, we get

$ \Rightarrow {x^{m + n}} \times {x^{n + 1}} \times {x^{1 + m}}$

Here, we can see that the base of the power is the same and exponents are different. So now by using the multiplication property of the powers, we get the equation as

$ \Rightarrow {x^{m + n}} \times {x^{n + 1}} \times {x^{1 + m}} = {x^{m + n + n + 1 + 1 + m}}$

And on solving the right side of the equation we will get the equation as

$ \Rightarrow {x^{m + n}} \times {x^{n + 1}} \times {x^{1 + m}} = {x^{2\left( {1 + m + n} \right)}}$ , and we will name it equation $1$

Now we will solve the denominator of the LHS side of the equation, so on solving it we will get

$ \Rightarrow {\left( {{x^m} \times {x^n} \times {x^1}} \right)^2} = {\left( {{x^{m + n + 1}}} \right)^2}$

And on solving furthermore, we will get the above equation as

$ \Rightarrow {x^{2\left( {m + n + 1} \right)}}$ , and we will name its equation $2$ .

So on dividing the equation $1$ with the equation $2$ , we get the equation as

$ \Rightarrow \dfrac{{{x^{2\left( {m + n + 1} \right)}}}}{{{x^{2\left( {m + n + 1} \right)}}}}$

And on solving the above equation, we can see there is a like term in both the numerator and denominator so on solving it we get

$ \Rightarrow 1$

Therefore, from this, we can see that LHS is equal to the RHS.