How do you simplify $ \dfrac{8}{\sqrt{3}} $ ?

Hint: We are asked to simplify $ \dfrac{8}{\sqrt{3}} $ we will first learn what are the things we need to simplify. We will learn about canceling terms, then will go for canceling radicals in the denominator as our term $ \dfrac{8}{\sqrt{3}} $ has $ \sqrt{3} $ radical in the denominator. So we use that cancellation process to simplify in this, we multiply and divide terms by $ \sqrt{3} $ and then the term we get after multiplication will be the required term.

Complete step by step answer:
We are given $ \dfrac{8}{\sqrt{3}} $ . We are asked to simplify it.
We are given a fraction in which the numerator is 8 and the denominator is $ \sqrt{3} $.
Before we solve it we must learn about how we need to solve such kind of problem.
If we are given a problem in a fraction like $ \dfrac{a}{b} $ we must first check that numerator a and denominator b have something in common or not, if they have something common then we can cancel those factors.
For example, consider $ \dfrac{18}{4} $ . We have numerator as 18 and denominator as 4.
We can see that, numerator and denominator has 2 as common because $ 18=2\times 9\text{ and }4=2\times 2 $ so, we first cancel that 2 so we get $ \dfrac{18}{4}=\dfrac{2\times 9}{2\times 2}=\dfrac{9}{2} $ .
Now after simplifying $ \dfrac{18}{4} $ we get the fraction as $ \dfrac{9}{2} $ .
We can see that our term $ \dfrac{8}{\sqrt{3}} $ has numerator 8 and denominator $ \sqrt{3} $ , the term has nothing in common, so we are good till here.
The second thing to check is that the denominator must never have a radical expression like $ \sqrt{b} $.
If we have number like $ \sqrt{b} $ in denominator, then we multiply and divide the term by $ \sqrt{b} $ and simplify it, like if we have term as $ \dfrac{a}{\sqrt{b}} $ so we multiply by $ \sqrt{b} $ and divide by $ \sqrt{b} $ we get, $ \dfrac{a}{\sqrt{b}}=\dfrac{a\times \sqrt{b}}{\sqrt{b}\times \sqrt{b}}=\dfrac{a\sqrt{b}}{b} $ .
Now we can see that our fraction $ \dfrac{8}{\sqrt{3}} $ has a radical in the denominator, so we will multiply and divide it by $ \sqrt{3} $ . So we get $ \dfrac{8}{\sqrt{3}}=\dfrac{8\times \sqrt{3}}{\sqrt{3}\times \sqrt{3}} $ .
As we know $ \sqrt{3}\times \sqrt{3}=3 $ so we get $ \dfrac{8\sqrt{3}}{3} $ .
Hence we get our term $ \dfrac{8}{\sqrt{3}} $ as $ \dfrac{8\sqrt{3}}{3} $ after simplification.

Note:
 If we have two terms in denominator with radical like $ \dfrac{a}{b+\sqrt{c}} $ then we will multiply and divide it by $ b-\sqrt{c} $ so we will have $ \dfrac{a}{b+\sqrt{c}}=\dfrac{a}{b+\sqrt{c}}\times \dfrac{b-\sqrt{c}}{b-\sqrt{c}} $ .
Simplifying we get $ \dfrac{ab-a\sqrt{c}}{{{b}^{2}}-c} $ .
This will be our simplified form.