Simplify Radical Expressions: A Step-by-Step Guide

Hey guys! Let's dive into simplifying a fascinating radical expression. This article will break down the steps to solve this mathematical puzzle, making it super easy to understand. We'll go through each term, rationalize the denominators, and combine like terms. So, buckle up and let's get started!

Understanding the Problem

Before we jump into the solution, let's understand what we're dealing with. We have the expression:

$$\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}+\frac{3 \sqrt{2}}{\sqrt{6}+\sqrt{3}}-\frac{4 \sqrt{3}}{\sqrt{6}+\sqrt{2}}$$

Our mission is to simplify this expression into its most basic form. This involves rationalizing denominators and combining terms. Sounds like a plan? Let's break it down step by step.

Step 1: Rationalizing the Denominators

Rationalizing the denominators is the first crucial step in simplifying this expression. To rationalize a denominator, we need to get rid of the square roots in the denominator. We achieve this by multiplying the numerator and denominator by the conjugate of the denominator. Remember, the conjugate of a binomial expression $a + b$ is $a - b$, and vice versa. This trick helps us eliminate the square roots in the denominator by using the difference of squares formula: $(a + b)(a - b) = a^2 - b^2$.

Term 1: $\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}$

Let's start with the first term: $\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}$. The conjugate of the denominator $\sqrt{2}+\sqrt{3}$ is $\sqrt{2}-\sqrt{3}$. So, we multiply both the numerator and the denominator by this conjugate:

$$\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}} \times \frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}} = \frac{\sqrt{6}(\sqrt{2}-\sqrt{3})}{(\sqrt{2}+\sqrt{3})(\sqrt{2}-\sqrt{3})}$$

Now, let’s expand the numerator and the denominator. In the numerator, we have $\sqrt{6}(\sqrt{2}-\sqrt{3}) = \sqrt{6} \times \sqrt{2} - \sqrt{6} \times \sqrt{3} = \sqrt{12} - \sqrt{18}$. Simplifying further, $\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$ and $\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}$. So, the numerator becomes $2\sqrt{3} - 3\sqrt{2}$.

For the denominator, we use the difference of squares formula: $(\sqrt{2}+\sqrt{3})(\sqrt{2}-\sqrt{3}) = (\sqrt{2})^2 - (\sqrt{3})^2 = 2 - 3 = -1$. Thus, our first term simplifies to:

$$\frac{2\sqrt{3} - 3\sqrt{2}}{-1} = 3\sqrt{2} - 2\sqrt{3}$$

Term 2: $\frac{3 \sqrt{2}}{\sqrt{6}+\sqrt{3}}$

Moving on to the second term: $\frac{3 \sqrt{2}}{\sqrt{6}+\sqrt{3}}$. The conjugate of $\sqrt{6}+\sqrt{3}$ is $\sqrt{6}-\sqrt{3}$. We multiply both the numerator and the denominator by this conjugate:

$$\frac{3 \sqrt{2}}{\sqrt{6}+\sqrt{3}} \times \frac{\sqrt{6}-\sqrt{3}}{\sqrt{6}-\sqrt{3}} = \frac{3 \sqrt{2}(\sqrt{6}-\sqrt{3})}{(\sqrt{6}+\sqrt{3})(\sqrt{6}-\sqrt{3})}$$

Expanding the numerator, we get $3\sqrt{2}(\sqrt{6}-\sqrt{3}) = 3\sqrt{2} \times \sqrt{6} - 3\sqrt{2} \times \sqrt{3} = 3\sqrt{12} - 3\sqrt{6}$. Simplifying, $3\sqrt{12} = 3\sqrt{4 \times 3} = 3 \times 2\sqrt{3} = 6\sqrt{3}$. So, the numerator becomes $6\sqrt{3} - 3\sqrt{6}$.

For the denominator, we again use the difference of squares formula: $(\sqrt{6}+\sqrt{3})(\sqrt{6}-\sqrt{3}) = (\sqrt{6})^2 - (\sqrt{3})^2 = 6 - 3 = 3$. Therefore, the second term simplifies to:

$$\frac{6\sqrt{3} - 3\sqrt{6}}{3} = 2\sqrt{3} - \sqrt{6}$$

Term 3: $\frac{4 \sqrt{3}}{\sqrt{6}+\sqrt{2}}$

Lastly, let's rationalize the denominator of the third term: $\frac{4 \sqrt{3}}{\sqrt{6}+\sqrt{2}}$. The conjugate of $\sqrt{6}+\sqrt{2}$ is $\sqrt{6}-\sqrt{2}$. We multiply both the numerator and the denominator by this conjugate:

$$\frac{4 \sqrt{3}}{\sqrt{6}+\sqrt{2}} \times \frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}} = \frac{4 \sqrt{3}(\sqrt{6}-\sqrt{2})}{(\sqrt{6}+\sqrt{2})(\sqrt{6}-\sqrt{2})}$$

Expanding the numerator, we have $4\sqrt{3}(\sqrt{6}-\sqrt{2}) = 4\sqrt{3} \times \sqrt{6} - 4\sqrt{3} \times \sqrt{2} = 4\sqrt{18} - 4\sqrt{6}$. Simplifying, $4\sqrt{18} = 4\sqrt{9 \times 2} = 4 \times 3\sqrt{2} = 12\sqrt{2}$. Thus, the numerator becomes $12\sqrt{2} - 4\sqrt{6}$.

Using the difference of squares formula for the denominator, we get $(\sqrt{6}+\sqrt{2})(\sqrt{6}-\sqrt{2}) = (\sqrt{6})^2 - (\sqrt{2})^2 = 6 - 2 = 4$. Hence, the third term simplifies to:

$$\frac{12\sqrt{2} - 4\sqrt{6}}{4} = 3\sqrt{2} - \sqrt{6}$$

Step 2: Combining the Simplified Terms

Now that we've rationalized all the denominators, we can rewrite the original expression with our simplified terms:

$$3\sqrt{2} - 2\sqrt{3} + 2\sqrt{3} - \sqrt{6} - (3\sqrt{2} - \sqrt{6})$$

Let's simplify this further by removing the parentheses and combining like terms:

$$3\sqrt{2} - 2\sqrt{3} + 2\sqrt{3} - \sqrt{6} - 3\sqrt{2} + \sqrt{6}$$

Notice anything cool? We have $3\sqrt{2}$ and $-3\sqrt{2}$, which cancel each other out. Also, we have $-2\sqrt{3}$ and $+2\sqrt{3}$, which also cancel out. And guess what? $-\sqrt{6}$ and $+\sqrt{6}$ cancel each other out too!

So, after all the cancellations, we are left with:

$$0$$

Conclusion

Guys, we did it! The simplified form of the given expression is 0. It might seem surprising that such a complex-looking expression boils down to zero, but that’s the beauty of mathematics. By systematically rationalizing the denominators and combining like terms, we were able to unravel the puzzle.

Remember, the key to simplifying radical expressions lies in rationalizing the denominators and combining like terms. Practice makes perfect, so keep at it, and you'll become a pro in no time!

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