Simplify: \[\dfrac{{{{\left( {25} \right)}^{\dfrac{3}{2}}} \times {{\left( {343} \right)}^{\dfrac{3}{5}}}}}{{{{\left( {16} \right)}^{\dfrac{5}{4}}} \times {{\left( 8 \right)}^{\dfrac{4}{3}}} \times {{\left( 7 \right)}^{\dfrac{3}{5}}}}}\]

Hint: For the simplification of the given expression, we can write the numbers in their powers of one of the prime numbers constituting them, according to the need. Then we can add and subtract the powers of the same base when multiplied and divided respectively to obtain the required answer.

Complete step-by-step answer:
We have been given the following expression for simplification:
\[\dfrac{{{{\left( {25} \right)}^{\dfrac{3}{2}}} \times {{\left( {343} \right)}^{\dfrac{3}{5}}}}}{{{{\left( {16} \right)}^{\dfrac{5}{4}}} \times {{\left( 8 \right)}^{\dfrac{4}{3}}} \times {{\left( 7 \right)}^{\dfrac{3}{5}}}}}\]
This expression can be simplified by writing these numbers in the power of their prime factor. So the powers to which these numbers are raised gets cancelled.
The number of times that factor appears will be its power.
The numbers in the expressions can be written as:
 $
  25 \to 5 \times 5 = {\left( 5 \right)^2} \\
  343 \to 7 \times 7 \times 7 = {\left( 7 \right)^3} \\
  16 \to 2 \times 2 \times 2 \times 2 = {\left( 2 \right)^4} \\
  8 \to 2 \times 2 \times 2 = {\left( 2 \right)^3} \\
  $
Substituting these values in the given expression, we get:\to
\[\dfrac{{{{\left( {{5^2}} \right)}^{\dfrac{3}{2}}} \times {{\left( {{7^3}} \right)}^{\dfrac{3}{5}}}}}{{{{\left( {{2^4}} \right)}^{\dfrac{5}{4}}} \times {{\left( {{2^3}} \right)}^{\dfrac{4}{3}}} \times {{\left( 7 \right)}^{\dfrac{3}{5}}}}}\]
When the same base numbers are multiplied, their powers are added and when they are divided, their powers are subtracted. Using this, we get:
$\Rightarrow \dfrac{\left( 5 \right)^3 \times \left( 7 \right)^{\dfrac{9}{5} - \dfrac{3}{5}} }{\left( 2 \right)^{5+4}}$
\[ \Rightarrow \left[ {\dfrac{{\left( {{5^3}} \right)}}{{\left( {{2^9}} \right)}}} \right]{7^{\dfrac{6}{5}}} \\
   \Rightarrow \left( {\dfrac{{125}}{{512}}} \right){7^{\dfrac{6}{5}}} \\
 \]
Therefore, the required value for the given expression is \[\left( {\dfrac{{125}}{{512}}} \right){7^{\dfrac{6}{5}}}\]
So, the correct answer is “\[\left( {\dfrac{{125}}{{512}}} \right){7^{\dfrac{6}{5}}}\]”.

Note: 343 could also have been written in the powers of prime number 3, but there would have been no other number alike. 343 is written in the powers of 7 due to the presence of its like term in the denominator.
While writing the numbers in the powers of their primes, we have to keep in mind that our main motive is to remove the denominator of the fractional power. So, there are 3 reasons that 16 was written in the powers of 2 and not 4 that are:
i) We prefer the prime numbers and 4 is not one but 2 is.
ii) Presence of 8 which is 2 raised to the power 3, it became the like term
iii) The denominator of fractional power was 4 which get cancelled when 16 was written in powers of 2 which would not have been possible if written in power of 4