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gorica
Group Title
I have few problems:
1) x+y+z=0
x^2+y^2+z^2=2
x^4+y^4+z^4=?
2) x+y+z=a
x^2+y^2+z^2=3a^2/2
x^3+y^3+z^3=a^3
xy+yz+xz=?
xyz=?
x^4+y^4+z^4=?
3) x/a + y/b + z/c = 1
a/x + b/y + c/z = 0
x^2/a^2 + y^2/b^2 + z^2/c^2 = ?
4) (7+5*(2^(1/2)))^(1/3)+(75*(2^(1/2)))^(1/3)
 one year ago
 one year ago
gorica Group Title
I have few problems: 1) x+y+z=0 x^2+y^2+z^2=2 x^4+y^4+z^4=? 2) x+y+z=a x^2+y^2+z^2=3a^2/2 x^3+y^3+z^3=a^3 xy+yz+xz=? xyz=? x^4+y^4+z^4=? 3) x/a + y/b + z/c = 1 a/x + b/y + c/z = 0 x^2/a^2 + y^2/b^2 + z^2/c^2 = ? 4) (7+5*(2^(1/2)))^(1/3)+(75*(2^(1/2)))^(1/3)
 one year ago
 one year ago

This Question is Open

gorica Group TitleBest ResponseYou've already chosen the best response.0
4) \[\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{75\sqrt{2}}\]
 one year ago

philo1234 Group TitleBest ResponseYou've already chosen the best response.0
R u there? If so let's look at each problem.
 one year ago

philo1234 Group TitleBest ResponseYou've already chosen the best response.0
What is the topic that you got these questions from?
 one year ago

gorica Group TitleBest ResponseYou've already chosen the best response.0
powers and roots
 one year ago

philo1234 Group TitleBest ResponseYou've already chosen the best response.0
What does the directions for these questions say?
 one year ago

gorica Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

philo1234 Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

gorica Group TitleBest ResponseYou've already chosen the best response.0
sure, if u find solution  post it :)
 one year ago

philo1234 Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

philo1234 Group TitleBest ResponseYou've already chosen the best response.0
Put these solution in the equation you are trying to find and solve from there. #1: Answer=2
 one year ago

gorica Group TitleBest ResponseYou've already chosen the best response.0
but I have to write how did I get it o.O
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.0
Since, x^2+y^2+z^2=2 1<=x<=1 , 1<=y<=1 and 1<=z<=1
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.0
From, x+y+z=0 u will get one of them is 1 , the other one is 1 and the other must be 0
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.0
So, x^4+y^4+z^4=2
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.0
\[\large xy+yz+xz=\frac{ 1 }{ 4 }a^2\]\[\large xyz=\frac{1}{4}a^3\]using algebra
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.0
3) x/a + y/b + z/c = 1 a/x + b/y + c/z = 0 x^2/a^2 + y^2/b^2 + z^2/c^2 =1
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.0
Let x/a=i , y/b=j , z/c=k then, i+j+k=1 1/i + 1/j+1/k=0 jk+ik+ij=0 Now, (i+j+k)^2=i^2+j^2+k^2+2(ij+ik+jk) (1)^2=i^2+j^2+k^2+2*0 Thus, i^2+j^2+k^2=1 x^2/a^2 + y^2/b^2 + z^2/c^2 =1
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.1
LOL this looks familiar. For the first one, except for saurav's logical reasoning, you can go for this: \((x+y+z)^2=0\) \(x^2+y^2+z^2 + 2(xy+xz+yz)=0\) \(2 + 2(xy+xz+yz)=0\) \(xy+xz+yz=1\) \((x^2+y^2+z^2)^2=x^4+y^4+z^4 + 2(x^2y^2+x^2z^2+z^2y^2)\) \((xy+xz+yz)^2=1\) \(x^2y^2+x^2z^2+z^2y^2 + 2(xyz)(x+y+z)=1\) \(x^2y^2+x^2z^2+z^2y^2 =1\) so,\((2)^2=x^4+y^4+z^4 + 2(1)\) \(x^4+y^4+z^4 =2\)
 one year ago

gorica Group TitleBest ResponseYou've already chosen the best response.0
thank you, guys :)
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.1
You're welcome :)
 one year ago
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