# Math Help - Confusing System of Equations

1. ## Confusing System of Equations

Hi MHF

Before I continue I would like to ask you guys for some help.
I want to know where/how can I practice hard Algebra questions?

How did you guys get so good? Any tips? Any book? Any website?
Thanks

Now, for the question

We have to find the value of $x^4+y^4+z^4$ when $x$ , $y$ and $z$ are real numbers.

Following three equalities:

$x+y+z=3$
$x^2+y^2+z^2=9$
$xyz$ $=-2$

First, from the first two equations we have (this is actually in the question)
$xy+yz+xz=A$

Next using

$(x^2+y^2+z^2)^2=x^4+y^4+z^4+B[(xy)^2+(yz)^2+(zx)^2]$

We have
$(x^4)+(y^4)+(z^4)=C$

I know NOTHING about this question. It asks to find $A,B$ and $C$ ;all these equations are given.

So,
any tips?

Last time I asked something here I got a really good response and actually got to know that I had missed somethings in school.
Am I missing out something?

Thanks a lot!!!

2. To find out A, B and C, we have to use the three equalities.

To find A, expand the first equality. Substitute using the other 2 equalities when necessary.

To find B, expand (x^2+y^2+z^2)^2. Rearrange the expanded equation and you can get B eventually.

From the question, we know that x^4+y^4+z^4 = C. So, we can rewrite the equation (x^2+y^2+z^2)^2 = C + B[(xy)^2+(yz)^2+(zx)^2)]. And the tricky part now is to find [(xy)^2+(yz)^2+(zx)^2)] which is by expanding the equation (xy+yz+xz)^2. Substitute when necessary and you will get the answer C.

Sorry that my explaination might be difficult to understand. Do ask if you have more questions.

Thanks.

I really thought there was an easy way to do this, like some property from the systems of equations that I had forgotten.
This requires a lot of work, doesn't it? hahaha

I really appreciate it :3

Edit:

So, I was trying to get A in this question, and it's not that simple.
Because x² is actually 9-3y-3z-3y+y²+yz-3z+yz+z² (using the FOIL method)
and not only x²=9-y²-z² (expanding the second equation)

I got to go in circles, what one equation does the other undoes.

This so...er... head-aching! I'm sorry, sometimes I really suck in algebra.

4. Hello, Zellator!

I have a different approach.
I'll get you started . . .

$\text{Given: }\;\begin{array}{ccccc}x+y+z &=& 3 & [1] \\ x^2+y^2+z^2 &=& 9 & [2] \\ xyz &=& \text{-}2 & [3] \end{array}$

$\text{Find: }\;x^4 + y^4 + z^4$

$\text{Square [1]: }\qquad\qquad\; (x + y + z)^2 \;=\;3^2$

. . $x^2 + 2xy + 2xz + y^2 + 2yz + z^2 \;=\;9$

. . $\underbrace{x^2 + y^2 + z^2}_{\text{This is 9}} +\;2(xy + yz + xz) \;=\;9$

. . . . . . . .Hence: . $\boxed{xy + yz + xz \;=\;0}$

$\text{Cube [1]: }\qquad (x + y + z)^3 \;=\;3^3$

$x^3 + y^3 + z^3 + 3x^2y + 3xy^2 + 3y^zz + 3yz^2 + 3x^2z + 3xz^2 + 6xyz \;=\;27$

$x^3 + y^3 + z^3 + 3x^2y + 3xy^2 + 3y^zz + 3yz^2 + 3x^2z + 3xz^2 + {\bf9xyz} \;=\;27 + {\bf3xyz}$

$x^3 + y^3 + z^2 + 3(x^2y + xy^2 + xyz) + 3(xyz + y^2z + yz^2) + 3(x^2z + xyz + xz^2)$
. . . . . . $=\;27 + 3xyz$

$x^3 + y^3 + z^2 + 3xy(x + y + z) + 3yz(x + y + z) + 3xz(x + y + z) \;=\;27 + 3xyz$

$x^3 + y^3 + z^3 + 3\underbrace{(x+y+z)}_{\text{This is 3}}\underbrace{(xy + yz + xz)}_{\text{This is 0}} \;=\;27+ 3\underbrace{(xyz)}_{\text{This is -2}}$

. . . . . . . . . . . . . . . $x^3 + y^3 + z^3 + 3(3)(0) \;=\;27 - 6$

. . . . . . . . . . . . . . . . . . . . . $\boxed{x^3 + y^3 + z^3 \;=\;21}$

Get the idea?

5. Absolutely! Wow!

I'm loss for words!