- How to prove $2x^2 + 2y^2 + 2z^2 -2xy -2yz \\geq 0$
I know how to prove $2x^2 + 2y^2 + 2z^2 -2xy -2yz -2xz \geq 0$ since I can split it up in the three terms $(x-y)^2, (y-z)^2, (z-x)^2$ which are all greater than or equal to zero, but how do I prove this without the last term: $2x^2 + 2y^2 + 2z^2 -2xy -2yz \geq 0$?
- partial differential equations - Solve the PDE $(x-y)p+(x+y)q=2xz . . .
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- Integer solutions of the equation $x^2+y^2+z^2 = 2xyz$
There are no solutions The original equation, considered by Markov, was $$ x^2 + y^2 + z^2 = 3xyz $$ This leads to the Markov Numbers
- How to Factorise $x^2 + y^2 + z^2 - 2xy - 2yz - 2zx$
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- linear algebra - Surface described by the equation $-3y^2 - 4xy + 2xz . . .
Given the equation : $-3y^2 - 4xy + 2xz + 4yz - 2x - 2z + 1 = 0$ Check if the surface described by that equation has a center of symmetry and then by making the correct coordinate system change, find the type of the surface that this equation describes
- $1$st Order PDE - Mathematics Stack Exchange
I was solving 1st order PDE which was \begin{equation} (x^2-y^2-z^2)p+2xyq=2xz \tag{1} \end{equation} I had tried to solve this
- Classify the surface $x^2 + y^2 - z^2 + 2xy - 2xz - 2yz - y = 0$
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- linear algebra - Writing an expression as a sum of squares . . .
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