Spooky Quadratic Substitution

[Ebba Sen Pai]

It may not be Halloween, but this problem still spooks me into request for assistance.

\[ x^6-10x^3+16=0 \]

I know I can substitute \[ x^6 \] for \[ y^2 \] and \[ x^3 \] for y.

​

From here I can solve with relative ease until I get to "y= 8, and 2". I re-substitute back my original variables and their powers so I have \[ x^3=8, x^3=2 \]
I can solve for x^3=8 by subtracting both sides by 8, to get "x^3-8=0", and factoring which provides me a difference of cubes and ultimately the correct solutions.
What spooks me beyond belief is trying factor "x^3-2=0", which I do not think can be factored. It certainly won't fit the difference of cubes as far as I can tell. I have no memory of being taught this in my book and have went back multiple times trying to understand what to do with \[ x^3-2= 0 \]
If anyone can help explain what should be done, or refer me to information I would be humbled.

 

[pka]

It may not be Halloween, but this problem still spooks me into request for assistance.
\[ x^6-10x^3+16=0 \]

I know I can substitute \[ x^6 \] for \[ y^2 \] and \[ x^3 \] for y. From here I can solve with relative ease until I get to "y= 8, and 2". I re-substitute back my original variables and their powers so I have \[ x^3=8, x^3=2 \]​

I can solve for x^3=8 by subtracting both sides by 8, to get "x^3-8=0", and factoring which provides me a difference of cubes and ultimately the correct solutions.
What spooks me beyond belief is trying factor "x^3-2=0", which I do not think can be factored. It certainly won't fit the difference of cubes as far as I can tell. I have no memory of being taught this in my book and have went back multiple times trying to understand what to do with \[ x^3-2= 0 \]
If anyone can help explain what should be done, or refer me to information I would be humbled.

Just do it: \(\displaystyle x^6-10x^3+16=(x^3-8)(x^3-2)=0\)So \(\displaystyle (x^3-8)=0\) OR \(\displaystyle (x^3-2)=0\) thus [\(\displaystyle (x=\sqrt[3]8\) OR \(\displaystyle (x=\sqrt[3]2\)
Those are the only real solutions, SEE HERE
If you know complex variables you can find four more. For each of \(\displaystyle 2~\&~8\) has three cube roots.

 

[ksdhart2]

Both \(\displaystyle x^3 = 8\) and \(\displaystyle x^3 = 2\) can be solved in exactly the same way. The only catch is that the latter problem doesn't have a nice, whole number solution. The key here is to think about what exponents mean. When you're solving \(\displaystyle x^3 = 8\), what you're really asking yourself is "what number, when multiplied by itself 3 times, equals 8?" A little bit of trial and error, or simply memorizing powers of some whole numbers tells you the answer must be 2.

So now what about \(\displaystyle x^3 = 2\). Well, again, we want to know "what number, when multiplied by itself 3 times, equals 8?" Immediately, we can tell that the number must be less than 2, since any number greater than 1 only ever increases in value when it's multiplied by itself. Similarly, since any number less than 1 only ever decreases in value when multiplied by itself, and 1 stays the same value when multiplied by itself, we also know that the number in question must be bigger than 1. So we have: \(\displaystyle 1 < x < 2\). But can we do better? Sure, I suppose with a little trial and error and calculator use, we can see that \(\displaystyle 1.5^3 = 3.375\) is too big, \(\displaystyle 1.25^3 = 1.953125\) is close but still too small, but then \(\displaystyle 1.26^3 = 2.000376\) is almost exact but just a hair too big. Clearly, honing in on the exact number will be very tedious. There must be a better way!

Indeed, there is. What we've discovered is that not only is the exact answer not a whole number, its decimal expansion actually continues forever without repeating (we call such numbers irrational). And this is where the aforementioned concept of exponents come into play. With exponents, we have a nice, compact "symbol" that serves as a stand-in for this exact value. Let's head on over to Math is Fun and see what they have to say about exponents and the laws associated therewith.

I see two laws that seems promising on that page: \(\displaystyle (x^{m})^{n} = x^{mn}\) and \(\displaystyle x^{\frac{1}{n}} = \sqrt[n]{x}\). What if we combined those two to solve this problem? Starting from the given, we have:

\(\displaystyle x^3 = 2 \implies (x^3)^{\frac{1}{3}} = 2^{\frac{1}{3}} \implies x^{3 \cdot \frac{1}{3}} = 2^{\frac{1}{3}} \implies x = \sqrt[3]{2}\)

 

[Steven G]

It may not be Halloween, but this problem still spooks me into request for assistance.

\[ x^6-10x^3+16=0 \]

I know I can substitute \[ x^6 \] for \[ y^2 \] and \[ x^3 \] for y.

​

From here I can solve with relative ease until I get to "y= 8, and 2". I re-substitute back my original variables and their powers so I have \[ x^3=8, x^3=2 \]
I can solve for x^3=8 by subtracting both sides by 8, to get "x^3-8=0", and factoring which provides me a difference of cubes and ultimately the correct solutions.
What spooks me beyond belief is trying factor "x^3-2=0", which I do not think can be factored. It certainly won't fit the difference of cubes as far as I can tell. I have no memory of being taught this in my book and have went back multiple times trying to understand what to do with \[ x^3-2= 0 \]
If anyone can help explain what should be done, or refer me to information I would be humbled.

Although you do not need to use this here it is: 2 = (2^1/3)³. So you can write x³-2 as the difference of two cubes as x³-(2^1/3)³.