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What is the topic that you got these questions from?
powers and roots
What does the directions for these questions say?
no directions. just what I wrote here. in first 3 questions, I have some expressions given and I have to evaluate those with question mark. The 4th one I just have to calculate. I know that I can use Lagrange's identity for sqrt, but I don't know what to do with 3th root.
Sorry wish I could help. This is the 1st time I have come across these type of questions. I will keep looking into them if I am able to find the solutions I will try to post it, if you are still stuck.
sure, if u find solution - post it :)
I found the solution for 1 and 2:
#1: x=-1,y=1,z=0 or change it around as u please.
#2: x=-a/2, y=a/2 and z=a or change it around as u see fit.
Put these solution in the equation you are trying to find and solve from there.
but I have to write how did I get it o.O
-1<=x<=1 , -1<=y<=1 and -1<=z<=1
u will get one of them is -1 , the other one is 1 and the other must be 0
LOL this looks familiar. For the first one, except for saurav's logical reasoning, you can go for this:
\(x^2+y^2+z^2 + 2(xy+xz+yz)=0\)
\(2 + 2(xy+xz+yz)=0\)
\((x^2+y^2+z^2)^2=x^4+y^4+z^4 + 2(x^2y^2+x^2z^2+z^2y^2)\)
\(x^2y^2+x^2z^2+z^2y^2 + 2(xyz)(x+y+z)=1\)
so,\((2)^2=x^4+y^4+z^4 + 2(1)\)