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Simplifying radicals in practice

Solving exercises about working with radicals, rationalizing the denominator.
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Hello. Let us try to solve the first exercise of the simplifying of radicals in practice step. The first exercise asks us to rationalize this expression. What means to rationalize an expression means that we want to rewrite the same thing, but without roots in the denominator. How can we do that? Of course, it would be great to have the possibility to take the cube of this quantity and the cube of this quantity. Let us see how to realize our dream. Remember, there is very important and notable identity, which is the following. That a cubed plus b cubed is equal to a plus b times a squared minus a times b plus b squared. Why this identity is useful for our aim?
94.3
Because look, here we have the sum of the cubic root of 4 plus cubic root of 2. And if we multiply this quantity by– look, a now is cubic root of 4, and b is the cubic root of 2. Then if we multiply this quantity by the square of the cubic root of 4 minus the cubic root of 4 times the cubic root of 2 plus the square of the cubic root of 2, what do we get? Exactly as it is written here, we get the cube of the cubic root of 4 plus the cube of the cubic root of 2, which is exactly 4 times 2. And you see we succeeded in eliminating the roots.
176.7
How can we apply this technique to our situation? Look, we consider 6 over the cubic root of 4 plus the cubic root of 2. And we multiply this by 1 in such a way of course we get again the quantity that we had in advance. But now we will write 1 in a very special way. 6 over cubic root of 4 plus cubic root of 2 times what? And we write now a fraction with the same numerator and denominator in such a way here we have again 1, but which will be the denominator?
232.7
Exactly this quantity, which is the cubic root of 4 square is the cubic root of 16 minus the cubic root of 8 plus the square of the cubic root of 2 is the cubic root of 4. And also at the numerator, we write, again, the same expression.
274.2
Then this fraction is 1. And now what happens? Observe. At the denominator, we have exactly this product, that is 4 plus 2, 6. And in the numerator, we have 6 times the cubic root of 16 minus the cubic root of 8 is 2 plus the cubic root of 4.
320.5
Then we can just eliminate the 6. And to arrive at the final result, which is the cubic root of 16 minus 2 plus the cubic root of 4. You see, we have realized our dream. We have rewritten this same expression in a new form. And in this new form, we have not any more roots at the denominator. In this case, we have no denominator at all, but of course, it will be not always the case. Thank you for your attention.

The following exercises are solved in this step.

We invite you to try to solve them before watching the video.

In any case, you will find below a PDF file with the solutions.

Exercise 1.

Rationalise the following expression: [ frac{6}{sqrt[3]{4}+sqrt[3]{2}}.]

Exercise 2. [Solved only in the PDF file]

Rationalise the following expression: [ frac{sqrt[3]{x}-y}{sqrt{y}+sqrt[6]{x}}.]

Exercise 3. [Solved only in the PDF file]

Rationalise the following expression: [ frac{2x-sqrt[3]{4x^2}}{sqrt{2x}-sqrt[3]{2x}}.]

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Precalculus: the Mathematics of Numbers, Functions and Equations

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