Expand the given expression ${{(3a\text{ }+\text{ }4b)}^{3}}$

Hint: Here we will use the formula of expansion of cubes, then we replace the terms of formula of expansion of cubes with the terms given in the question and then we proceed with the solution by using the BODMAS rule so as to avoid mistakes.

Complete step-by-step answer:
We know the formula of expansion of cubes which is
${{(x\text{ }+\text{ }y)}^{3}}\text{ }=\text{ }{{x}^{3}}\text{ }+{{y}^{3}}\text{ }+\text{ }3xy(x\text{ }+\text{ }y)$ ………………$\left( 1 \right)$
So here we will replace x with 3a and y with 4b.
Therefore,
$x\text{ }=\text{ }3a$
$y\text{ }=\text{ }4b$
Now the expansion becomes;
 ${{(3a\text{ }+\text{ }4b)}^{3}}\text{ }=\text{ }{{(3a)}^{3}}\text{ }+{{(4b)}^{3}}\text{ }+\text{ }3\times 3a\times 4b(3a+\text{ }4b)$
$=\text{ }27{{a}^{3}}\text{ }+64{{b}^{3}}\text{ }+\text{ }3\times 3a\times 4b\times 3a+3\times 3a\times 4b\times 4b$
$=\text{ }27{{a}^{3}}\text{ }+64{{b}^{3}}\text{ }+\text{ }108{{a}^{2}}b+144a{{b}^{2}}$
So, the expansion of ${{(3a\text{ }+\text{ }4b)}^{3}}$is $27{{a}^{3}}\text{ }+64{{b}^{3}}\text{ }+\text{ }108{{a}^{2}}b+144a{{b}^{2}}$.

Additional information: There are some other commonly used formulas of cubes which are sum of cubes and difference of cubes:-
 ${{x}^{3}}-{{y}^{3}}=\left( x-y \right)\left( {{x}^{2}}+xy+{{y}^{2}} \right)$ and ${{x}^{3}}+{{y}^{3}}=\left( x+y \right)\left( {{x}^{2}}-xy+{{y}^{2}} \right)$

Note: Formula for the expansion of the cube of difference of the terms can be derived easily by replacing the second term with a negative of the given second term.
Example: we can write ${{(x\text{ }-\text{ }y)}^{3}}$as ${{(x+\text{ }(-\text{ }y))}^{3}}$, now we can use formula in equation$\left( 1 \right)$ to calculate its expansion by only replacing y with –y in formula. After calculation, we get the final expansion of ${{(x\text{ }-\text{ }y)}^{3}}$as ${{x}^{3}}\text{ }-\text{ }{{y}^{3}}\text{ }-\text{ }3xy(x\text{ }-\text{ }y)$.