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gorica
 4 years ago
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)
gorica
 4 years ago
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)

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gorica
 4 years ago
Best ResponseYou've already chosen the best response.04) \[\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{75\sqrt{2}}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0R u there? If so let's look at each problem.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What is the topic that you got these questions from?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What does the directions for these questions say?

gorica
 4 years ago
Best ResponseYou've already chosen the best response.0no 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry 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.

gorica
 4 years ago
Best ResponseYou've already chosen the best response.0sure, if u find solution  post it :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Put these solution in the equation you are trying to find and solve from there. #1: Answer=2

gorica
 4 years ago
Best ResponseYou've already chosen the best response.0but I have to write how did I get it o.O

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Since, x^2+y^2+z^2=2 1<=x<=1 , 1<=y<=1 and 1<=z<=1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0From, x+y+z=0 u will get one of them is 1 , the other one is 1 and the other must be 0

sirm3d
 4 years ago
Best 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.03) 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Let 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0LOL 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,\((2)^2=x^4+y^4+z^4 + 2(1)\) \(x^4+y^4+z^4 =2\)
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