MHB Solving $x$ in Terms of $a$: $x^4-3ax^3+3a^3x+a^4=0$

[anemone]

Solve for $x$ in terms of $a$ in the equation $x^4-3ax^3+3a^3x+a^4=0$.

 

[topsquark]

Great. Now I'm not getting any sleep tonight.

-Dan

Spoiler

That wasn't as bad as I thought. A little guess work and serendipity go a long way.

Let's start with the (almost preposterous) idea that it can be factored:
$$x^4 - 3ax^3 + 3a^3x + a^4 = (x^2 + bx + c)(x^2 + dx + e)$$

Expanding out gives use the equations:
$$b + d = -3a$$
$$c^2 + bd + e^2 = 0$$
$$be + cd = 3a^3$$
$$ce = a^4$$

Let's try $$c = e = a^2$$.

Then we get, from the first and third equations that $$b + d = -3a$$ and $$b + d = 3a$$. No good.

Let's try again. Let $$c = e = -a^2$$.

Then the first and third equations both read $$b + d = -3a$$. So we now have $$d = -3a - b$$.This means our last equation (the second one) now gives
$$b^2 + 3ab + 2a^2 = (b + 2a)(b + a) = 0$$

So we have that b = -2a or b = -a.

Both of them give the same answer, so let's put b = -2a.

Tallying up:
$$b = -2a$$
$$c = -a^2$$
$$d = -a$$
$$e = -a^2$$.

Thus (whew!)
$$x^4 - 3ax^3 + 3a^3x + a^4 = (x^2 - 2ax - a^2)(x^2 - ax - a^2) = 0$$

This solves down to the following four solutions:
$$x = a ( 1 \pm \sqrt{2} )$$
$$x = \frac{a}{2} ( 1 \pm \sqrt{5} )$$

-Dan

 

[anemone]

Great. Now I'm not getting any sleep tonight.

-Dan

Hi Dan,

I take this as you like this challenge problem and I wish you the best of luck to crack it it no time! (Sun)

 

[topsquark]

Hi Dan,

I take this as you like this challenge problem and I wish you the best of luck to crack it it no time! (Sun)

Actually, I did. :) A little motivation helps...I have to be up by 6.

-Dan

 

[kaliprasad]

my solution

Spoiler

X^4 – 3ax^3 + 3a^3 x + a^4 = 0
Forgetting the sign it is symmetric from left to right and a is increasing as x is decresing
So divide by x^2 and collect x^n and 1/x^n together

(X^2 + a^4/x^2) – 3a(x - a^2 /x) = 0
Or (x-a^2/x)^2 + 2a^2 – 3a(x-a^2/x) = 0
Putting x- a^2/x = y we get y^2 + 2a^2 – 3ay = 0
So y = a or 2a
So x^2 - 2ax – a^2 = 0 or x^2 –ax – a^2 = 0
both quadratic can be solved

 

[anemone]

my solution

Spoiler

X^4 – 3ax^3 + 3a^3 x + a^4 = 0
Forgetting the sign it is symmetric from left to right and a is increasing as x is decresing
So divide by x^2 and collect x^n and 1/x^n together

(X^2 + a^4/x^2) – 3a(x - a^2 /x) = 0
Or (x-a^2/x)^2 + 2a^2 – 3a(x-a^2/x) = 0
Putting x- a^2/x = y we get y^2 + 2a^2 – 3ay = 0
So y = a or 2a
So x^2 - 2ax – a^2 = 0 or x^2 –ax – a^2 = 0
both quadratic can be solved

Thanks for participating, kaliprasad...

But I don't quite agree with the end result that you obtained.

Take for example if $$x^2 - 2ax – a^2 = 0$$ is true, solving it for $x$ we get:

$$x=a\pm a\sqrt{2}=a(1\pm \sqrt{2})$$

Substitute the expression of $x=a(1+\sqrt{2})$ above into the original given equation, we have:

$$x^4 – 3ax^3 + 3a^3 x + a^4=(a(1+ \sqrt{2}))^4 – 3a(a(1+ \sqrt{2}))^3 + 3a^3(a(1+ \sqrt{2})+ a^4=-12a^4 \ne 0$$

What do you think, kaliprasad

 

[kaliprasad]

Thanks for participating, kaliprasad...

But I don't quite agree with the end result that you obtained.

Take for example if $$x^2 - 2ax – a^2 = 0$$ is true, solving it for $x$ we get:

$$x=a\pm a\sqrt{2}=a(1\pm \sqrt{2})$$

Substitute the expression of $x=a(1+\sqrt{2})$ above into the original given equation, we have:

$$x^4 – 3ax^3 + 3a^3 x + a^4=(a(1+ \sqrt{2}))^4 – 3a(a(1+ \sqrt{2}))^3 + 3a^3(a(1+ \sqrt{2})+ a^4=-12a^4 \ne 0$$

What do you think, kaliprasad

I did not check you calculation.
my solution matches with that of topsquark and further
(x^2-2ax - a^2)(x^2-ax - a^2) gives me the product

(x^4 - 3ax^3 + 3a^3 x +a4)where is the mistake

 

[anemone]

I'm so so sorry!

First, I didn't see topsquark has edited his post and added the solution in his first post...and second, I want to apologize for saying $x=a(1+\sqrt{2})$ isn't the solution to the given equation because it's...

And thanks for participating in this problem to both of you!:)