Evaluatethe expression $\dfrac{\sqrt[3]{729}-\sqrt[3]{27}}{\sqrt[3]{512}+\sqrt[3]{343}}$

Hint: Here, we need to find the cube root of the numbers 729, 27, 512, and 343 separately and then substitute the obtained values in the question and solve it further to find the required result.

Complete step-by-step solution:
In this question, here we need to solve the given question $\dfrac{\sqrt[3]{729}-\sqrt[3]{27}}{\sqrt[3]{512}+\sqrt[3]{343}}$, first let us find the cube root of 729, 27, 512 and 343.
For number 729:
Let us first find the factors of the number 729,
$729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 $
Here we can see that there are three pairs of the prime factor 3, but we need to find the cube root, so we will group three prime factors together, so we get
$729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 $
$\sqrt[3]{729}$ = $\sqrt[3]{3\times 3\times 3\times 3\times 3\times 3}$
             $= 3 \times 3 $
             $= 9$
Therefore, the cube root of the number 729 is 9
For number 27:
$27 = 3 \times 3 \times 3$
Since, it is a cube root, we need to group three prime factors of the same number.
$ 27 = 3 \times 3 \times 3$
$\sqrt[3]{27}$ = $\sqrt[3]{3\times 3\times 3}$
           = 3
Therefore, the cube root of the number 27 is 3.
For number 512:
$512 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$
Since, it is a cube root, we need to group three prime factors of the same number.
$512 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$
$\sqrt[3]{512}$ = $\sqrt[3]{2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2}$
             $= 2 \times 2 \times 2$
             $= 8$
Therefore, the cube root of 512 is 8.
For number 343:
$343 = 7 \times 7 \times 7$
Since, it is a cube root, we need to group three prime factors of the same number.
$343 = 7 \times 7 \times 7$
$\sqrt[3]{343}$ = $\sqrt[3]{7\times 7\times 7}$
             = 7
Therefore, the cube root of 343 is 7.
Let us write after we got the cube roots of the numbers 729, 27, 512 and 343 as 9, 3, 8 and 7 respectively.
$\dfrac{\sqrt[3]{729}-\sqrt[3]{27}}{\sqrt[3]{512}+\sqrt[3]{343}}$ = $\dfrac{9-3}{8+7}$
                             = $\dfrac{6}{15}$
                             = $\dfrac{2}{5}$
Hence, after evaluating we got the value as $\dfrac{2}{5}$.

Note: This particular question, can also be solved by multiplying the numerator and the denominator by the conjugate of the denominator. Conjugate of a number means where you change the sign, which is from ‘+’ to ‘- ‘or from ‘- ‘to ‘+’. If we solved by taking the conjugate of the denominator and multiply it in numerator and denominator, we would be able to solve the question but we would take much more time using the quadratic formulas and expansion which is not efficient.