Simplify the values :
(i). ${\left( {{1^3} + {2^3} + {3^3}} \right)^{\dfrac{1}{2}}}$
(ii). ${\left( {\dfrac{{81}}{{16}}} \right)^{\dfrac{3}{4}}} \times {\left( {\dfrac{{25}}{9}} \right)^{\dfrac{{ - 3}}{2}}}$

Hint: In order to simplify the equation, for the first equation solve the brackets by taking out the cube and adding them, then finding the square root of the obtained result. For the second equation initiate with solving the operands separately and finding the product of the two operands.

Complete step-by-step solution:
1. ${\left( {{1^3} + {2^3} + {3^3}} \right)^{\dfrac{1}{2}}}$
For the above equation solve the values inside the brackets, by taking cube of the operands as:
${1^3} = 1$, ${2^3} = 8$ and ${3^3} = 27$
Substituting the values in the equation:
${\left( {{1^3} + {2^3} + {3^3}} \right)^{\dfrac{1}{2}}} = {\left( {1 + 8 + 27} \right)^{\dfrac{1}{2}}}$
Solving the values inside the bracket:
${\left( {1 + 8 + 27} \right)^{\dfrac{1}{2}}}$
$ \Rightarrow {\left( {36} \right)^{\dfrac{1}{2}}}$
Since, we know that we can write $36$as the square of 6, so numerically it can be written as:
$ \Rightarrow {\left( {36} \right)^{\dfrac{1}{2}}} = {\left( {{6^2}} \right)^{\dfrac{1}{2}}}$
From Law of radicals, we know that ${\left( {{a^p}} \right)^q} = {a^{pq}}$, so accordingly we can write:
$ \Rightarrow {\left( {{6^2}} \right)^{\dfrac{1}{2}}} = {6^{2 \times \dfrac{1}{2}}}$
Solving the powers:
$ \Rightarrow {6^{2 \times \dfrac{1}{2}}} = {6^1} = 6$
Therefore, the value of ${\left( {{1^3} + {2^3} + {3^3}} \right)^{\dfrac{1}{2}}} = 6$.
2). ${\left( {\dfrac{{81}}{{16}}} \right)^{\dfrac{3}{4}}} \times {\left( {\dfrac{{25}}{9}} \right)^{\dfrac{{ - 3}}{2}}}$
For the above equation, we will write the value of the operands in the brackets in the form of powers.
We can write the values as follows:
$81 = {3^4}$, $16 = {2^4}$, $25 = {5^2}$ and $9 = {3^2}$.
Substituting all the values obtained above in the equation ${\left( {\dfrac{{81}}{{16}}} \right)^{\dfrac{3}{4}}} \times {\left( {\dfrac{{25}}{9}} \right)^{\dfrac{{ - 3}}{2}}}$, and we get:
${\left( {\dfrac{{81}}{{16}}} \right)^{\dfrac{3}{4}}} \times {\left( {\dfrac{{25}}{9}} \right)^{\dfrac{{ - 3}}{2}}} = {\left( {\dfrac{{{3^4}}}{{{2^4}}}} \right)^{\dfrac{3}{4}}} \times {\left( {\dfrac{{{5^2}}}{{{3^2}}}} \right)^{\dfrac{{ - 3}}{2}}}$
Taking 2 common from the numerator and denominator:
$ \Rightarrow {\left( {\dfrac{3}{2}} \right)^{4 \times \dfrac{3}{4}}} \times {\left( {\dfrac{5}{3}} \right)^{2 \times \dfrac{{ - 3}}{2}}}$
On solving the powers, we get:
$ \Rightarrow {\left( {\dfrac{3}{2}} \right)^3} \times {\left( {\dfrac{5}{3}} \right)^{ - 3}}$
Since, we know that ${\left( {\dfrac{a}{c}} \right)^{ - q}}$ can be written as $\left( {\dfrac{c}{a}} \right)$. So, accordingly we write ${\left( {\dfrac{5}{3}} \right)^{ - 3}} = {\left( {\dfrac{3}{5}} \right)^3}$. Substituting this value in the above equation:
$ \Rightarrow {\left( {\dfrac{3}{2}} \right)^3} \times {\left( {\dfrac{3}{5}} \right)^3}$
Taking the powers as common:
$ \Rightarrow {\left( {\dfrac{3}{2} \times \dfrac{3}{5}} \right)^3}$
$ \Rightarrow {\left( {\dfrac{9}{{10}}} \right)^3}$
Taking cube of the value:
$ \Rightarrow \dfrac{{{9^3}}}{{{{10}^3}}} = \dfrac{{9 \times 9 \times 9}}{{10 \times 10 \times 10}} = \dfrac{{729}}{{1000}}$
Therefore, the value of ${\left( {\dfrac{{81}}{{16}}} \right)^{\dfrac{3}{4}}} \times {\left( {\dfrac{{25}}{9}} \right)^{\dfrac{{ - 3}}{2}}} = \dfrac{{729}}{{1000}}$.

Note: If two operands in multiplication have their powers same, then their powers can be taken common as ${a^b} \times {c^b} = {\left( {ac} \right)^b}$. Remember the above statement is always true for operands in products or division. This case is not possible for sum or difference. For example: ${a^b} + {c^b} \ne {\left( {a + c} \right)^b}$.