:confused:I am having trouble with an algebraic problem which I have tried doing for a long time, now it is very close to the due date and I still don't have the answer... :(
The question is:
If x^2 + y^2 + z^2 + 10x + 20y - 30z + 350 = 0, find the value of (x-y-z)^5
If anyone is smart enough to do it, I will thank you a lot.
Very desperate.
Thank you

Complete the square.

(x^{2}+10x+25)+(y^{2}+20y+100)+(z^{2}-30z+225)=-350+25+100+125=0

Factor:

(x+5)^{2}+(y+10)^{2}+(z-15)^{2}=0

Now, the values of x,y,z are right in front of you. Find those, then plug them into

(x-y-z)^{5} and you're done.

Complete the square.

Factor:

(x+5)^{2}+(y+10)^{2}+(z-15)^{2}=0

Now, the values of x,y,z are right in front of you. Find those, then plug them into

(x-y-z)^{5} and you're done.

What i don't get is how (x+5)^{2}+(y+10)^{2}+(z-15)^{2}=0 can change to (x-y-z)^{5}.
I've done the first few steps and i get how you get to the factor but I'm a little confused about how the equations with ^{2} can turn into ^{5} .
Thanks for you help, I have appreciated it a lot :) :p

It does not change to (x-y-z)^{5}.

Simply find the values in the equation I showed after completing the square and factoring, then plug them into that.

What makes (x+5)^{2}=0?

What makes (y+10)^{2}=0?

What makes (z-15)^{2}=0?

Easy enough. Find those and plug them in.

I think you're just making this a little harder than need be. See what I mean now?

Any number squared will be greater than zero (unless you are including complex numbers).

So, if (X+5)^2 + (Y+10)^2 + (Z-15)^2 = 0

then

each of the terms must also equal zero (X+5)^2 = 0. There is no other way to add three non-negative numbers and come up with zero.

Solve those three equations for X, Y, and Z and you'll have your answer.