Math Problem help

**gfbrd** · November 4th 2012, 07:08 PM

Hey there sorry if this is in the wrong thread, but I need some help with this problem

If x^2+y^2+z^2=49, and x+y+z=x^3+y^3+z^3=7
find xyz

I've been trying this problem for a while now and was hoping if someone can guide me through this problem so I can understand it better thanks.

**Soroban** · November 5th 2012, 06:49 AM

Hello, gfbrd!

I'm familiar with this type of problem,
but this one took me a couple of tries . . .

$\text{Given: }\:\begin{Bmatrix}x+y+z \:=\:7 & [1] \\ x^2+y^2+z^2 \:=\:49 & [2] \\ x^3+y^3 +z^3 \:=\:7 & [3] \end{Bmatrix}$

$\text{Find }xyz.$

Square [1]: . $(x+y+z)^2 \:=\:7^2$

. . $x^2 + 2xy + 2xz + y^2 + 2xz + z^2 \:=\:49$

. . $2(xy + yz + xz) + \underbrace{(x^2+y^2+z^2)}_{\text{This is 49}} \:=\:49$

Hence: . $2(xy+yz+xz) \:=\:0 \quad\Rightarrow\quad xy + yz + xz \:=\:0\;\;[4]$

Cube [1]: . $(x+y)^3 \:=\:7^3$

. . $\begin{bmatrix}x^3 \\ 3x^2y + 3x^2z \\ 3xy^2 + 6xyz + 3xz^2 \\ y^3 + 3y^2z + 3yz^2 + z^3 \end{bmatrix} \;=\;343$

Add $3xyz$ to both sides:

. . $\begin{bmatrix}x^3 \\ 3x^2y + 3x^2z \\ 3xy^2 + 9xyz + 3xz^2 \\ y^3 + 3y^2z + 3yz^2 + z^3 \end{bmatrix} \;=\;343 + 3xyz$

The left side is:

. . $\overbrace{(x^3 + y^3 + z^3)}^{\text{This is 7}} + (3x^2y + 3x^2y + 3xyz) + (3y^2z + 3yz^2 + 3xyz)$
. . . . $+ (3x^2z + 3xz^2 + 3xyz) \;=\; 343 + 3xyz$

. . $7 + 3xy(x+y+z) + 3yz(x+y+z) + 3xz(x+y+z) \;=\;343 + 3xyz$

. . $7 + 3\underbrace{(xy + yz + xz)}_{\text{This is 0}}\underbrace{(x+y+z)}_{\text{This is 7}} \;=\;343 + 3xyz$

Therefore: . $7 \;=\;343 + 3xyz \quad\Rightarrow\quad 3xyz \:=\:-336 \quad\Rightarrow\quad \boxed{xyz \:=\:-112}$

**gfbrd** · November 5th 2012, 07:50 AM

Nice work soroban, I was working on this for so long and had trouble with it, thank you so much.

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