Solve the expression: $\dfrac{{{{\left( {25} \right)}^{\dfrac{3}{2}}} \times {{\left( {243} \right)}^{\dfrac{3}{5}}}}}{{{{\left( {16} \right)}^{\dfrac{5}{4}}} \times {{\left( 8 \right)}^{\dfrac{4}{3}}}}}$.

Hint: In these types of questions firstly we have to check if there is a square, cube or any other power of any number present or not. Then we can easily simplify this question. But for that, we must know the square or any other power of some numbers. As in this question, we can easily find that there are some numbers whose roots exist.

Complete step-by-step solution:
In the given question,
$ = \dfrac{{{{\left( {25} \right)}^{\dfrac{3}{2}}} \times {{\left( {243} \right)}^{\dfrac{3}{5}}}}}{{{{\left( {16} \right)}^{\dfrac{5}{4}}} \times {{\left( 8 \right)}^{\dfrac{4}{3}}}}}$
Now,
We can also write,
${\left( 5 \right)^2} = 25$
${\left( 3 \right)^5} = 243$
${\left( 2 \right)^4} = 16$
${\left( 2 \right)^3} = 8$
Therefore,
$ = \dfrac{{{{\left( {{{\left( 5 \right)}^2}} \right)}^{\dfrac{3}{2}}} \times {{\left( {{{\left( 3 \right)}^5}} \right)}^{\dfrac{3}{5}}}}}{{{{\left( {{{\left( 2 \right)}^4}} \right)}^{\dfrac{5}{4}}} \times {{\left( {{2^3}} \right)}^{\dfrac{4}{3}}}}}$
Now, multiplying the powers
$ = \dfrac{{{{\left( 5 \right)}^{2 \times \dfrac{3}{2}}} \times {{\left( 3 \right)}^{5 \times \dfrac{3}{5}}}}}{{{{\left( 2 \right)}^{4 \times \dfrac{5}{4}}} \times {{\left( 2 \right)}^{3 \times \dfrac{4}{3}}}}}$
On calculation, we get
$ = \dfrac{{{{\left( 5 \right)}^3} \times {{\left( 3 \right)}^3}}}{{{{\left( 2 \right)}^5} \times {{\left( 2 \right)}^4}}}$
As we know that, ${\left( 5 \right)^3} = 125\,\,,\,\,{\left( 3 \right)^3} = 27\,\,,\,\,{\left( 2 \right)^5} = 32\,\,and\,\,{\left( 2 \right)^4} = 16$
$ = \dfrac{{125 \times 27}}{{32 \times 16}}$
On multiplication, we get
$ = \dfrac{{3375}}{{512}}$
Hence, the required value is $\dfrac{{3375}}{{512}}$.

Additional information: The square root of a negative is not a real number. There is no real number multiplied by itself that will equal to a negative number. For example, $\sqrt { - 25} $ does not equal $ - 5$
because $ - 5$ times $ - 5$ is not $ - 25$ , but $25$ .Algebraically speaking, for real numbers $\sqrt {{a^2}} $ to equal a, a must not be negative. It can be $0$ because the square root of ${0^2}$ or $0$ times $0$ is $0$.

Note: The root of a number x is another number, which when multiplied by itself a given number of times, equals x. This would be spoken as “the third root of $64$ is $4$ “or “the cube root of $64$ is $4$”.
The second root is called the “square root”. There are no real even-order roots of negative numbers. For example, there is no real square root of $ - 9$ , because $ - 3 \times 3 = + 9$ and $ + 3 \times + 3 = 9$ also.
This applies to all even-order roots, $2nd$ (square) root, $4th$ root, $6th$ root, and so on.