How do you simplify $2$ cubed root $81$$ + $$3$ cubed root $24$?

Hint: Here, we are asked to solve $2\sqrt[3]{{81}} + 3\sqrt[3]{{24}}$ To solve this , we first need to look at the values whose cube roots are involved in the problem and determine whether they are perfect cubes or whether they contain factors which are perfect cubes. Then we will find the cube roots of these perfect cubes and simplify the expression.

Complete step-by-step solution:
We need to simplify the term $2\sqrt[3]{{81}} + 3\sqrt[3]{{24}}$. We will first consider the two numbers given in the cube root and convert them as the product of numbers one of which will be the perfect cube. The first number is 81. This number can be written as a product of two numbers 3 and 27. We know that 27 is the perfect cube, so we can easily find its cube root.

We can write
$81 = 3 \times 27 = 3 \times 3 \times 3 \times 3 = 3 \times {3^3}$
The second number is 24. This number can be written as a product of two numbers 3 and 8. We know that 8 is the perfect cube, so we can easily find its cube root.
We can write
$24 = 3 \times 8 = 3 \times 2 \times 2 \times 2 = 3 \times {2^3}$
Now, we will simplify the given expression.
$
  2\sqrt[3]{{81}} + 3\sqrt[3]{{24}} \\
   = 2\sqrt[3]{{3 \times {3^3}}} + 3\sqrt[3]{{3 \times {2^3}}} \\
 $
We know that $\sqrt[3]{{{a^3}}} = a$
Therefore,
$
  2\sqrt[3]{{81}} + 3\sqrt[3]{{24}} \\
   = 2\sqrt[3]{{3 \times {3^3}}} + 3\sqrt[3]{{3 \times {2^3}}} \\
   = 2 \times 3\sqrt[3]{3} + 3 \times 2\sqrt[3]{3} \\
   = 6\sqrt[3]{3} + 6\sqrt[3]{3} \\
   = 12\sqrt[3]{3} \\
 $
Thus, our final answer is $12\sqrt[3]{3}$.

Note: Here, we have determined the cube roots while solving this question. To find the cubic root of a number easily, we can use the prime factorization method. By evaluating the prime factors we can pair similar digits in a group of three and take them out as a single digit from the cubic root.