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Rationalize the denominator and simplify:
\[\dfrac{{\sqrt 6 + \sqrt 3 }}{{\sqrt 6 - \sqrt 3 }}\]

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Last updated date: 13th Jun 2024
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Answer
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Hint: To solve the question, the first step that we will do is to rationalize the given expression. This involves multiplying both numerator and denominator by the same irrational number. This results in easy simplification further.

Complete step-by-step solution:
The given expression is:
\[\dfrac{{\sqrt 6 + \sqrt 3 }}{{\sqrt 6 - \sqrt 3 }}\]
Here we will use identity to rationalize the denominator of the given expression.
The identity is;
$(a+b)(a-b)={a^2} - {b^2}$.
Here the denominator is \[\sqrt 6 - \sqrt 3 \] which is of the form $(a-b)$. So, we will multiply both numerator and denominator by \[\sqrt 6 + \sqrt 3 \] to get the form of the above identity in the denominator.
So, on multiplying \[\sqrt 6 + \sqrt 3 \] with numerator and denominator we get,
\[\dfrac{{(\sqrt 6 + \sqrt 3 )(\sqrt 6 + \sqrt 3 )}}{{(\sqrt 6 - \sqrt 3 )(\sqrt 6 + \sqrt 3 )}} = \dfrac{{{{(\sqrt 6 + \sqrt 3 )}^2}}}{{{{(\sqrt 6 )}^2} - {{(\sqrt 3 )}^2}}}\]
Using the identity (a+b)(a-b)=${a^2} - {b^2}$ and ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$, the above expression can be simplified as:
\[\dfrac{{6 + 3 + 2\sqrt {18} }}{{6 - 3}} = \dfrac{{9 + 2\sqrt {18} }}{3} = \dfrac{{9 + 2\sqrt {9 \times 2} }}{3} = \dfrac{{3(3 + 2\sqrt 2 )}}{3} = 3 + 2\sqrt 2 \]
Therefore, the simplified expression is \[3 + 2\sqrt 2 \].

Note: You should know about rationalization. It is the process of eliminating a radical or imaginary number from the denominator of an algebraic function by multiplying the same factor in the numerator and denominator. You should know that $\sqrt a \times \sqrt b = \sqrt {ab}$ where a, b are positive numbers.