Can this be simplified? 1 / [3 * cube root of {x^2}]

[Phenomniverse]

I'm working on a problem and I have arrived at the following answer. I'm wondering if the denominator of the second term can be simplified somehow?

If you can't make it out, its 3 times the cube root of x squared.
Thanks.

 

[tkhunny]

I'm working on a problem and I have arrived at the following answer. I'm wondering if the denominator of the second term can be simplified somehow?

If you can't make it out, its 3 times the cube root of x squared.
Thanks.

You'll have to do better than that. What you have suggested doesn't seem to have a denominator. Please use your best efforts to communicate what it is you are doing.

 

[Phenomniverse]

I pasted an image of the question into my post, but somehow it didn't come out.
The term I'm trying to simplify (if possible) is 1 / (3 * cube root of x^2)

This is what I'm working with:

\(\displaystyle \large{=\, \dfrac{5}{2\, \sqrt{\strut x\,}}\, -\, \dfrac{1}{3\, \sqrt[3]{\strut x^2\,}}}\)

 

[sinx]

\(\displaystyle \large{=\, \dfrac{5}{2\, \sqrt{\strut x\,}}\, -\, \dfrac{1}{3\, \sqrt[3]{\strut x^2\,}}}\)

finding a common denominator first, then add (subtract) and simplify.
hint; write the powers and roots of x as exponents.

 

[sinx]

I pasted an image of the question into my post, but somehow it didn't come out.
The term I'm trying to simplify (if possible) is 1 / (3 * cube root of x^2)

how do you write cube root as an exponent?

 

[HallsofIvy]

Square root is the half power, cube root is the third power. The "common denominator" is 6. Write the two roots as 6th roots, then combine fractions.

 

[JeffM]

\(\displaystyle \large{=\, \dfrac{5}{2\, \sqrt{\strut x\,}}\, -\, \dfrac{1}{3\, \sqrt[3]{\strut x^2\,}}}\)

There is a technique called rationalizing a denominator, which is itself merely a special case of the technique of multiplying a numeral or expression by some form of 1 to get a new numeral or expression in a more convenient form but with the same numeric value.

Here we go:

$\displaystyle \dfrac{5}{2\sqrt{x}} - \dfrac{1}{3\sqrt[3]{x^2}} = \left ( 1 * \dfrac{5}{2\sqrt{x}} \right ) - \dfrac{1}{3\sqrt[3]{x^2}} = \left ( \dfrac{\sqrt{x}}{\sqrt{x}} * \dfrac{5}{2\sqrt{x}} \right ) - \dfrac{1}{3\sqrt[3]{x^2}} =$

$\displaystyle \dfrac{5\sqrt{x}}{2(\sqrt{x})^2} - \dfrac{1}{3\sqrt[3]{x^2}} = \dfrac{5\sqrt{x}}{2x} - \dfrac{1}{3\sqrt[3]{x^2}}.$

Do you see that we no longer have a radical in the denominator of the first fraction? Do you also see that the new expression represents the same quantity as the original expression?

You can multiply the second fraction by 1 without changing its value. But multiplying it by

$\displaystyle \dfrac{\sqrt{x}}{\sqrt{x}}$

will not eliminate the radical in the denominator of the second fraction. What will?

EDIT: The suggestions made in other posts will more quickly get you to the same result as my suggestion, but may be less intuitive to you.

 

[Phenomniverse]

Thanks for the suggestions.

So I guess if I can rationalise the second fraction, I can find a common factor of both denominators and then subtract the fractions.
But I don't have a strong intuitive sense of what a cube root of a square number is, so I'm struggling to figure out what version of '1' to multiply the second fraction by to rationalise its denominator.
I get that the cube root of x squared can be rewritten as x to the power of 2 thirds. And I think what this means is basically the number that when multiplied by itself three times will be equal to x squared?
Does that mean I need to multiply the denominator by (the cube root of x squared) squared, to end up with 3x^2 as the denominator, etc? If I carry on like this, I get a common denominator of 6x^2, and a big ugly numerator. See attached pic. Does it look right? Is this a better way of writing it than what I started with? Can the brackets in the second half of the numerator be expanded?

 

[sinx]

Thanks for the suggestions.

So I guess if I can rationalise the second fraction, I can find a common factor of both denominators and then subtract the fractions.
But I don't have a strong intuitive sense of what a cube root of a square number is, so I'm struggling to figure out what version of '1' to multiply the second fraction by to rationalise its denominator.
I get that the cube root of x squared can be rewritten as x to the power of 2 thirds. And I think what this means is basically the number that when multiplied by itself three times will be equal to x squared?
Does that mean I need to multiply the denominator by (the cube root of x squared) squared, to end up with 3x^2 as the denominator, etc? If I carry on like this, I get a common denominator of 6x^2, and a big ugly numerator. See attached pic. Does it look right? Is this a better way of writing it than what I started with? Can the brackets in the second half of the numerator be expanded?
View attachment 9994

sqrtx=x^1/2, cube rt x²=x^2/3
you want a common denominator.
so you need a common power of x, (and of course common number in front).
i.e. change 2x^1/2 and 3x^2/3 to 6x^2/6 , [and of course multiply top of each fraction by the req'd term].
then you can combine the fractions.

 

[Dr.Peterson]

Thanks for the suggestions.

So I guess if I can rationalise the second fraction, I can find a common factor of both denominators and then subtract the fractions.
But I don't have a strong intuitive sense of what a cube root of a square number is, so I'm struggling to figure out what version of '1' to multiply the second fraction by to rationalise its denominator.
I get that the cube root of x squared can be rewritten as x to the power of 2 thirds. And I think what this means is basically the number that when multiplied by itself three times will be equal to x squared?
Does that mean I need to multiply the denominator by (the cube root of x squared) squared, to end up with 3x^2 as the denominator, etc? If I carry on like this, I get a common denominator of 6x^2, and a big ugly numerator. See attached pic. Does it look right? Is this a better way of writing it than what I started with? Can the brackets in the second half of the numerator be expanded?
View attachment 9994

This is very good. There are a few places where you could make things a little simpler, but that comes with experience!

The one thing that needs to be done is to rewrite $\displaystyle \left ( \sqrt[3]{x^2} \right )^2$ as $\displaystyle \sqrt[3]{\left ( x^2 \right )^2} = \sqrt[3]{x^4}$.

The things you did that can be simpler are multiplying by $\displaystyle \left ( \sqrt[3]{x} \right )$ rather than $\displaystyle \left ( \sqrt[3]{x^2} \right )^2$ (try it, and see why it works), and using the LCD, $\displaystyle 6x^2$, rather than what I call the Obvious Common Denominator, $\displaystyle 2x \cdot 3x^2$, for the fractions.

 

[Phenomniverse]

This is very good. There are a few places where you could make things a little simpler, but that comes with experience!

The one thing that needs to be done is to rewrite $\displaystyle \left ( \sqrt[3]{x^2} \right )^2$ as $\displaystyle \sqrt[3]{\left ( x^2 \right )^2} = \sqrt[3]{x^4}$.

The things you did that can be simpler are multiplying by $\displaystyle \left ( \sqrt[3]{x} \right )$ rather than $\displaystyle \left ( \sqrt[3]{x^2} \right )^2$ (try it, and see why it works), and using the LCD, $\displaystyle 6x^2$, rather than what I call the Obvious Common Denominator, $\displaystyle 2x \cdot 3x^2$, for the fractions.

Ok that helps a lot. Thanks.
Just to clarify, moving the square from outside the brackets to under the root sign $\displaystyle \left ( \sqrt[3]{x^2} \right )^2$ as $\displaystyle \sqrt[3]{\left ( x^2 \right )^2}$ doesn't change the meaning, i.e that it is the whole term that is squared, not just the term under the root?

The suggestion to multiply by $\displaystyle \left ( \sqrt[3]{x} \right )$ rather than $\displaystyle \left ( \sqrt[3]{x^2} \right )^2$ streches my (limited) intuition about cube roots, but I'm thinking that if I write them as powers I get x to the 1/3 times x to the 2/3, which results in x to the 3/3 or just x??

 

[JeffM]

Thanks for the suggestions.

So I guess if I can rationalise the second fraction, I can find a common factor of both denominators and then subtract the fractions.
But I don't have a strong intuitive sense of what a cube root of a square number is, so I'm struggling to figure out what version of '1' to multiply the second fraction by to rationalise its denominator.
I get that the cube root of x squared can be rewritten as x to the power of 2 thirds. And I think what this means is basically the number that when multiplied by itself three times will be equal to x squared?
Does that mean I need to multiply the denominator by (the cube root of x squared) squared, to end up with 3x^2 as the denominator, etc? If I carry on like this, I get a common denominator of 6x^2, and a big ugly numerator. See attached pic. Does it look right? Is this a better way of writing it than what I started with? Can the brackets in the second half of the numerator be expanded?
View attachment 9994

First, you were going strong there.

Yes, $\displaystyle \sqrt[3]{x^2} * \sqrt[3]{x^2} * \sqrt[3]{x^2} = (\sqrt[3]{x^2})^2 = x^2.$

So to get rid of that cube root of x squared, you need to multiply by the square of the cube of x^2. (This is quicker to perceive using fractional exponents.)

$\displaystyle \dfrac{5\sqrt{x}}{2x} - \dfrac{1}{3\sqrt[3]{x^2}} = \dfrac{5\sqrt{x}}{2x} - \left ( \dfrac{1}{3\sqrt[3]{x^2}} * 1 \right) = \dfrac{5\sqrt{x}}{2x} - \dfrac{1}{3\sqrt[3]{x^2}} * \left ( \dfrac{\sqrt[3]{x^2}}{\sqrt[3]{x^2}} \right )^2 = \dfrac{5\sqrt{x}}{2x} - \dfrac{(\sqrt[3]{x^2})^2}{3x^2}.$

Now lets consider simplifying the numerator of the second fraction.

$\displaystyle (\sqrt[3]{x^2})^2 = \sqrt[3]{x^2} * \sqrt[3]{x^2} = \sqrt[3]{x^2 * x^2} =$

$\displaystyle \sqrt[3]{x^4} = \sqrt[3]{x^3 * x} = x\sqrt[3]{x}.$

$\displaystyle \therefore \dfrac{5\sqrt{x}}{2x} - \dfrac{(\sqrt[3]{x^2})^2}{3x^2} = \dfrac{5\sqrt{x}}{2x} - \dfrac{x\sqrt[3]{x}}{3x^2} = \dfrac{5\sqrt{x}}{2x} - \dfrac{\sqrt[3]{x}}{3x}.$

When you work with radicals, remember to simplify them as well.

And now the common denominator work is really easy.

$\displaystyle \dfrac{5\sqrt{x}}{2x} - \dfrac{\sqrt[3]{x}}{3x} = \dfrac{5\sqrt{x}}{2x} * \dfrac{3}{3} - \dfrac{\sqrt[3]{x}}{3x} * \dfrac{2}{2} = \dfrac{15\sqrt{x} - 2\sqrt[3]{x}}{6x}.$

Notice that the process of getting common denominators also involves multiplying by 1 in a convenient form.

Now check your work. 64 is a convenient number to work with because it is the smallest number greater than 1 that is both a perfect square and a perfect cube. (Never check a simplification with 0 or 1.)

$\displaystyle \dfrac{5}{2\sqrt{64}} - \dfrac{1}{3\sqrt[3]{64^2}} = \dfrac{5}{2 * 8} - \dfrac{1}{3 * 16} = \dfrac{5}{16} - \dfrac{1}{48} = \dfrac{15 - 1}{48} = \dfrac{14}{48} = \dfrac{7}{24}.$

$\displaystyle \dfrac{15\sqrt{64} - 2\sqrt[3]{64}}{6 * 64} = \dfrac{15 * 8 - 2 * 4}{6 * 64} =\dfrac{8(15 - 1)}{6 * 64} = \dfrac{14}{6 * 8} = \dfrac{7}{24}.$

So it checks.

Now as others have said, you can do this more quickly using fractional exponents so I'd make sure to become comfortable with them.

 

[JeffM]

Ok that helps a lot. Thanks.
Just to clarify, moving the square from outside the brackets to under the root sign $\displaystyle \left ( \sqrt[3]{x^2} \right )^2$ as $\displaystyle \sqrt[3]{\left ( x^2 \right )^2}$ doesn't change the meaning, i.e that it is the whole term that is squared, not just the term under the root?

The suggestion to multiply by $\displaystyle \left ( \sqrt[3]{x} \right )$ rather than $\displaystyle \left ( \sqrt[3]{x^2} \right )^2$ streches my (limited) intuition about cube roots, but I'm thinking that if I write them as powers I get x to the 1/3 times x to the 2/3, which results in x to the 3/3 or just x??

As Dr. Peterson says, you learn simplifying tricks with experience.

$\displaystyle \sqrt[3]{x^2} * \sqrt[3]{x} = \sqrt[3]{x^2 * x} = \sqrt[3]{x^3} = x.$

So that was a simpler way than you chose, but, as my previous post showed, your way would have got you to the correct answer if you had simplified a radical, which is something else experience teaches you.

And yes $\displaystyle x^{1/3} * x^{2/3} = x^{(1/3) + (2/3)} = x^1 = x.$

And lastly, you seem to be mystifying roots and squares and cubes. They are operations on numbers that result in numbers.

$\displaystyle \sqrt[3]{8^2} = \sqrt[3]{64} = 4.$ First we do one operation, then the other.

Now it is true that most roots are irrational numbers, but still they are just numbers.

 

[Phenomniverse]

Thanks again to everyone for the help on this one, I've learned a lot. I have another similar question, but I'll start another thread for that one.

 

[Dr.Peterson]

Ok that helps a lot. Thanks.
Just to clarify, moving the square from outside the brackets to under the root sign $\displaystyle \left ( \sqrt[3]{x^2} \right )^2$ as $\displaystyle \sqrt[3]{\left ( x^2 \right )^2}$ doesn't change the meaning, i.e that it is the whole term that is squared, not just the term under the root?

The suggestion to multiply by $\displaystyle \left ( \sqrt[3]{x} \right )$ rather than $\displaystyle \left ( \sqrt[3]{x^2} \right )^2$ streches my (limited) intuition about cube roots, but I'm thinking that if I write them as powers I get x to the 1/3 times x to the 2/3, which results in x to the 3/3 or just x??

I'll make one additional comment.

The change from $\displaystyle \left ( \sqrt[3]{x^2} \right )^2$ to $\displaystyle \sqrt[3]{\left ( x^2 \right )^2}$ does change the "meaning", in the sense that the first says to first take the cube root and then square, while the second says to first square and then take the cube root. But it doesn't change the value, because there is a property of roots and exponents that makes them equivalent.

As has been pointed out, this can most easily be seen in terms of fractional exponents. The first expression means $\displaystyle \left ( \left (x^2\right )^{1/3} \right )^2$, while the second means $\displaystyle \left ( \left (x^2 \right )^2\right )^{1/3}$, and both are equivalent to $\displaystyle x^{2 \cdot 2 \cdot 1/3} = x^{4/3}$.

 

[Denis]

You can assign a value to x, then evaluate the initial expression;
to see if the resulting expressions that you calculate are correct,
simply evaluate them "as you go"...get my drift?