Question

CHALLENGE QUESTION
Solve in positive real number the system of equation :

\((2x)^{2013} + (2y)^{2013} + (2z)^{2013} = 3\)

and \(xy + yz + zx + 2xyz = 1\)

Answer 1

Enjoy \(2013\) :)

Answer 2

Nothing to do here.

Answer 3

;)

Answer 4

@ParthKohli nothing to do?

Answer 5

x = y = z = 1/2

proving :
1^3 + 1^3 + 1^3 = 3 true
1/4 + 1/4 + 1/4 + 1/4 = 1 true too
hohohoho :)

Answer 6

:)

Answer 7

setting up the lagrangian should be enough, wouldn't it?

Answer 8

I think it would involve "modulus" but I don't know how..

Answer 9

\[f(x,y,z)=x^{2013}+y^{2013}+z^{2013}-3\times2^{-2013}\\
g(x,y,z)=xy+yz+zx+2xyz\\\;\\
f_x=\lambda g_x\implies2013x^{2012}=\lambda(y+z+2yz)\\
f_y=\lambda g_y\implies2013y^{2012}=\lambda(x+z+2xz)\\
f_z=\lambda g_z\implies2013z^{2012}=\lambda(x+y+2xy)\\
g(x,y,z)=c\implies xy+yz+zx+2xyz=1
\]

Answer 10

That looks like you have tried something tex work but I would have to go through that first ...

Answer 11

from 1,3
\[y+z=\frac{2013x^{2012}}{\lambda}-2yz\\
x(y+z)+yz+2xyz=1\implies \frac{2013x^{2013}}{\lambda}-2xyz+yz+2xyz=1\\
\implies\boxed{\lambda yz=2013x^{2013}}
\]
similarly, others
\(\lambda zx,\lambda xy\)

Answer 12

Gud work @electrokid

So, is that your final answer

Answer 13

thats half the steps.. find out lambda

Answer 14

Ok so go on.........

Answer 15

or use the three relations and find out the equations for "x","y" and "z"
lambdas cancel out

Answer 16

what do you men go on? go ahead and plug em up back into the relations..

Answer 17

jk

Answer 18

I would after that.......

Gud work