Confusing System of Equations

**Zellator** · May 4th 2011, 05:26 PM

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!!!

**ahyan1234** · May 4th 2011, 07:51 PM

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.

**Zellator** · May 6th 2011, 04:07 AM

Thanks for replying ahyan1234!

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

This is great!!!!! Thank you so much for your reply!
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.

**Soroban** · May 6th 2011, 06:19 AM

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?

**Zellator** · May 6th 2011, 07:16 AM

Absolutely! Wow!

I'm loss for words!
I really wish I had your experience in this hahahhaha!

I'm sorry, but why did you made the Cubic of the expression?
You probably need the cubic to get the ^4 right?

This is really amazing! It seem so easy when you guys do it!
I'll probably dig the forum up for more equations and problems like this.

Thanks for you time ahyan1234 and Soroban.

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