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# Help

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Simplify $\frac{2\sqrt[3]9}{1 + \sqrt[3]3 + \sqrt[3]9}.$

$$\frac{2\sqrt[3]9}{1 + \sqrt[3]3 + \sqrt[3]9}$$

Guest Feb 3, 2018
edited by Melody  Feb 3, 2018

#3
+2248
+3

After thinking about this one for quite some time, I think that I have found the magical conjugate: $$3^\frac{1}{3}-1$$.

First of all, I will convert all the radicals to exponential form.

$$\frac{2*9^\frac{1}{3}}{1+3^\frac{1}{3}+9^\frac{1}{3}}$$

Now, let's use the magical conjugate.

 $$\frac{2*9^\frac{1}{3}}{1+3^\frac{1}{3}+9^\frac{1}{3}}*\frac{3^\frac{1}{3}-1}{3^\frac{1}{3}-1}$$ Let's simplify the denominator first just to prove that this conjugate is indeed as magical as I claim. $$\left(1+3^\frac{1}{3}+9^\frac{1}{3}\right)\left(3^\frac{1}{3}-1\right)$$ Let's do the distributing of every term. $$1*3^\frac{1}{3}-1*1+3^\frac{1}{3}*3^\frac{1}{3}-1*3^\frac{1}{3}+9^\frac{1}{3}*3^\frac{1}{3}-1*9^\frac{1}{3}$$ This looks pretty menacing. Let's just simplify this monstrosity somewhat. $$3^\frac{1}{3}-1+\left(3^\frac{1}{3}\right)^2-3^\frac{1}{3}+9^\frac{1}{3}*3^\frac{1}{3}-9^\frac{1}{3}$$ This still does not look very nice. There is some cancelling that can occur here. $$-1+3^\frac{2}{3}+\left(3^2\right)^\frac{1}{3}*3^\frac{1}{3}-\left(3^2\right)^\frac{1}{3}$$ It is wise here to make every exponential term in the same base. This will allow the computation to occur. Let's continue simplifying to see what will happen. $$-1+3^\frac{2}{3}*3^\frac{1}{3}-3^\frac{2}{3}$$ Look at that! More exponential terms are cancelling out. In fact, they are all going away! $$-1+3$$ $$2$$

That's a glorious realization to make, huh? The magical conjugate has transformed such a complex denominator into something extraordinarily simple. Let's calculate the numerator now.

 $$\left(2*9^\frac{1}{3}\right)\left(3^\frac{1}{3}-1\right)$$ Once again, convert all exponents to common bases and distribute. $$9^\frac{1}{3}=\left(3^2\right)^\frac{1}{3}=3^\frac{2}{3}\\ 2*3^\frac{2}{3}*3^\frac{1}{3}-1*2*3^\frac{2}{3}$$ Notice how the multiplication involves a common base, so you can add the exponents together. $$2*3-2*3^\frac{2}{3}$$ Let's bring back that denominator of 2! $$\frac{2*3-2*3^\frac{2}{3}}{2}$$ Of couse, if both terms in the numerator has a factor of 2, like here, then we can simplify further. $$3-3^\frac{2}{3}\\ 3-\sqrt[3]{3^2}\Rightarrow 3-\sqrt[3]{9}$$ Since the problem was given in radical form, it is generally considered good form to remain consistent and provide the answer in radical form, too.
TheXSquaredFactor  Feb 3, 2018
#1
+93679
+1

$$\dfrac{2\sqrt[3]9}{1 + \sqrt[3]3 + \sqrt[3]9}\\ =\dfrac{2*(3^{1/3})^2}{1 + (3^{1/3}) + (3^{1/3})^2}\\$$

there is not much you can do with this.

Melody  Feb 3, 2018
#2
0

Melody, CPhill, Hectictar: The above reduces to: 3 - 3^(2/3), but cannot get that far !! See if you can!

Guest Feb 3, 2018
#3
+2248
+3

After thinking about this one for quite some time, I think that I have found the magical conjugate: $$3^\frac{1}{3}-1$$.

First of all, I will convert all the radicals to exponential form.

$$\frac{2*9^\frac{1}{3}}{1+3^\frac{1}{3}+9^\frac{1}{3}}$$

Now, let's use the magical conjugate.

 $$\frac{2*9^\frac{1}{3}}{1+3^\frac{1}{3}+9^\frac{1}{3}}*\frac{3^\frac{1}{3}-1}{3^\frac{1}{3}-1}$$ Let's simplify the denominator first just to prove that this conjugate is indeed as magical as I claim. $$\left(1+3^\frac{1}{3}+9^\frac{1}{3}\right)\left(3^\frac{1}{3}-1\right)$$ Let's do the distributing of every term. $$1*3^\frac{1}{3}-1*1+3^\frac{1}{3}*3^\frac{1}{3}-1*3^\frac{1}{3}+9^\frac{1}{3}*3^\frac{1}{3}-1*9^\frac{1}{3}$$ This looks pretty menacing. Let's just simplify this monstrosity somewhat. $$3^\frac{1}{3}-1+\left(3^\frac{1}{3}\right)^2-3^\frac{1}{3}+9^\frac{1}{3}*3^\frac{1}{3}-9^\frac{1}{3}$$ This still does not look very nice. There is some cancelling that can occur here. $$-1+3^\frac{2}{3}+\left(3^2\right)^\frac{1}{3}*3^\frac{1}{3}-\left(3^2\right)^\frac{1}{3}$$ It is wise here to make every exponential term in the same base. This will allow the computation to occur. Let's continue simplifying to see what will happen. $$-1+3^\frac{2}{3}*3^\frac{1}{3}-3^\frac{2}{3}$$ Look at that! More exponential terms are cancelling out. In fact, they are all going away! $$-1+3$$ $$2$$

That's a glorious realization to make, huh? The magical conjugate has transformed such a complex denominator into something extraordinarily simple. Let's calculate the numerator now.

 $$\left(2*9^\frac{1}{3}\right)\left(3^\frac{1}{3}-1\right)$$ Once again, convert all exponents to common bases and distribute. $$9^\frac{1}{3}=\left(3^2\right)^\frac{1}{3}=3^\frac{2}{3}\\ 2*3^\frac{2}{3}*3^\frac{1}{3}-1*2*3^\frac{2}{3}$$ Notice how the multiplication involves a common base, so you can add the exponents together. $$2*3-2*3^\frac{2}{3}$$ Let's bring back that denominator of 2! $$\frac{2*3-2*3^\frac{2}{3}}{2}$$ Of couse, if both terms in the numerator has a factor of 2, like here, then we can simplify further. $$3-3^\frac{2}{3}\\ 3-\sqrt[3]{3^2}\Rightarrow 3-\sqrt[3]{9}$$ Since the problem was given in radical form, it is generally considered good form to remain consistent and provide the answer in radical form, too.
TheXSquaredFactor  Feb 3, 2018
#5
+7324
+4

Wow!! 👍

hectictar  Feb 3, 2018
#6
+93679
+2

That is excellent work x^2   BUT

I would like to know what thought processes led you to finding that conjugate.

(Becasue that it the bit that stumped the rest of us)

Melody  Feb 4, 2018
#7
+2248
+3

Well, trial and error appeared to be my friend here. If I was not aware that the original expression could be simplified further, I probably would have made the same conclusion as you did, Melody. It is considered improper to have radicals or fractional exponents in the denominator, so I knew that there was some way to simplify this. My method is really only relevant to this particular problem. You will see why.

Now, let's consider that extra bit on the end, the $$9^\frac{1}{3}=\left(3^2\right)^\frac{1}{3}=3^\frac{2}{3}=3^{\left(\frac{1}{3}\right)^2}$$. This means that the denominator can be written like $$1+3^\frac{1}{3}+3^{\left(\frac{1}{3}\right)^2}$$, and if I let $$x=3^\frac{1}{3}$$, then I can represent the denominator like $$x^2+x+1$$. I then made a stunning realization. This required some extraordinary observational skills.

 $$x^3-y^3=(x-y)(x^2+xy+y^2)$$ I am sure that you are familiar with this. It's the factorization of a difference of cubes. Let's set y equal to 1 and substitute. $$x^3-1=(x-1)(x^2+x+1)$$ Wait a second! Look at what one of the factors is. It's $$x^2+x+1$$, which is also the expression of the denominator. This tells me that  if I multiply the trinomial in the denominator by x-1, then I will be left with two terms. This is perfect! I have now found a way to manipulate the denominator into two terms. Let's keep going. Since $$x=3^\frac{1}{3}$$, let's just substitute it in. $$3^{\left(\frac{1}{3}\right)^3}-1=\left(3^\frac{1}{3}-1\right)\left(3^{\left(\frac{1}{3}\right)^2}+3^\frac{1}{3}+1\right)$$ Let's complete the simplification here. The process is quite straightforward when you are comfortable with the law of indices. $$3^{\left(\frac{1}{3}\right)^3}-1\\ 3^{\frac{1}{3}*3}-1\\ 3^1-1\\ 3-1\\ 2$$ Of course, notice what I multiplied the denominator by in the original problem. It is $$3^\frac{1}{3}-1$$.

So, no, the conjugate is not "magical." The only issue with this thought process is that this will not help you on most types of problems with multi-term denominators. This is just one particular case, and I happened to crack it.

TheXSquaredFactor  Feb 4, 2018
#8
+93679
+1

That is really cool.

Thanks   XSquaredFactor

Melody  Feb 7, 2018
#4
+2

Bravo, X^2 !!!!.

Guest Feb 3, 2018