Retarded Algebra

I've been working on this problem in each moment of my spare time for nearly 3 days straight. If I don't figure this shit out, I'll go crazy. I think there's something simple that I'm just not getting.

a + b + c = a^2 + b^2 + c^2 = a^3 + b^3 + c^3 = 3/2.

What is the value of a times b times c?

is this real problem? fuck god this is stupid but cuz u aksed i gues i shud say that a b and c have to be grater than 1 but less than 1.5 indipendently cuz if u square a desimal it gets smaler.

 butt fuck.... if u say a+B+C = 3/2, taht means each a b and c is 1/2. so 1/2 tiems 1/2 tiemes 1/2 is 1/8

 fuckin problem is tho is in da otha part of da equajin.... taht 1/2^2 is 1/4 and 1/4 X 3 is 1/64... so wtf???

 

One of a, b, or c is equal to 1/2.

The other two have the values (1/2)(2 +- √6).

Their product is -1/16.

>>3

How did you arrive at this answer?

>>3

Bullshit, 1/2 + (1/2)(2+√6) + (1/2)(2-√6) = 5/2, and their product is -1/4.

>>5

replace

(1/2)(2 +- √6)

with

(1/4)(2 +- √6) you nigger

(%i1) solve([a+b+c=3/2,a^2+b^2+c^2=3/2,a^3+b^3+c^3=3/2],[a,b,c]);
(%o1)
[
[a=(sqrt(6)+2)/4,b=-(sqrt(2)*sqrt(3)-2)/4,c=1/2],
[a=(sqrt(6)+2)/4,b=1/2,c=-(sqrt(2)*sqrt(3)-2)/4],
[a=-(sqrt(6)-2)/4,b=(sqrt(2)*sqrt(3)+2)/4,c=1/2],
[a=-(sqrt(6)-2)/4,b=1/2,c
=(sqrt(2)*sqrt(3)+2)/4],
[a=1/2,b=(sqrt(6)+2)/4,c=-(sqrt(2)*sqrt(3)-2)/4],
[a=1/2,b=-(sqrt(6)-2)/4,c=(sqrt(2)*sqrt(3)+2)/4]
]

>>7

doesn't tell us how to solve it

>>8
It looks like there are 3 equations with 3 unknowns.  It looks like it's a pain in the ass to solve for a, substitute, etc.  I only had the patience to verify the above solutions with maple ... I would hope there is an elegant solution.

Retarded Algebra is Retarded

Problems as that usually don't have an elegant solution. Most textbook examples are engineered to give an nice solution but if it's just some random equation it probably wont'.

Okay, OP here, I think I have it.

Since the sum of a b and c must be 3/2, they must average to 1/2. There are an infinite number of combinations that can result in this, but for the sake of simplicity, I assumed that one was 1/2, and the other two were 1/2 +- x.

If you square them, you're able to solve for x, which turns out to be sqrt(6)/4.

Sub for x in the original equation and multiply the terms to get (1/2)(1/2 + sqrt(6)/4)(1/2 - sqrt(6)/4), which, as was stated above, gives a product of -1/16.

Here it is in the way I should have shown it in the first place:

a = 1/2
b = 1/2 - x
c = 1/2 + x

1/2 + (1/2 - x) + (1/2 + x) = 3/2
1/4 + (1/4 - x + x^2) + (1/4 + x + x^2) = 3/2
2x^2 = 3/4
x = sqrt(6/16)
x = sqrt(6)/4

(1/2)(1/2 + sqrt(6)/4)(1/2 - sqrt(6)/4) = -1/16.

I have not yet checked to see if that value of x works when you cube the terms, mostly due to my own laziness. I'll report back if it turns out that this solution is incorrect.

well, from a geometric perspective, a+b+c=3/2 is a plane and a^2+b^2+c^2=3/2 is a sphere, so their intersection ought to be a circle, which should be simple enough to parameterize.  From there you should be just a few trig identities away from a solution.  Too lazy to do it right now.

Solving for the first two eq.s, I get a=1/2+sin(t)/sqrt(2)-cos(t)/sqrt(5),  b=1/2-sin(t)/sqrt(2)-cos(t)/sqrt(5),  c=1/2+2*cos(t)/sqrt(5);  Now we just need some brave soul to put this into a^3+b^3+c^3=3/2 and hopefully we'll get some points (I'm fairly certain that if there are solutions, there'll be at least 3 of them)

There is a solution of a very similar problem at http://blog.plover.com/math/symmetric-function.html

Basically, you can
(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac = 9/4

Subtracting a^2 + b^2 + c^2 from both sides gives

2ab + 2bc + 2ac = 3/4.     (*)

(a+b+c)^3 = a^3 + b^3 + c^3
          + 3a^2b + 3a^2c + 3b2^a + 3b^2c + 3c^2a + 3c^2b
       = 27/8

Subtracting a^3 + b^3 + c^3 from both sides gives

3a^2b + 3a^2c + 3b2^a + 3b^2c + 3c^2a + 3c^2b = 15/8    (**)

Now take (*)
and multiply it by a + b + c = 3/2, getting

2a^2b + 2a^2c + 2b2^a + 2b^2c + 2c^2a + 2c^2b + 6abc = 9/8

Multiply both sides by 3/2:

3a^2b + 3a^2c + 3b2^a + 3b^2c + 3c^2a + 3c^2b + 9abc = 27/16

Subtract (**) from this, leaving

9abc = -3/16

so

abc = -1/48.

It's quite possible that I fucked up the arithmetic somewhere, but the method I showed should work fine, as long as you do it carefully on a big sheet of paper.

Hope this helps.

>>15

Oh, I see where the fuckup is.  My expansion of (a+b+c)^3 is missing the 6abc term.