gorica
  • gorica
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)+(7-5*(2^(1/2)))^(1/3)
Pre-Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
gorica
  • gorica
4) \[\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}\]
anonymous
  • anonymous
R u there? If so let's look at each problem.
gorica
  • gorica
i am

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anonymous
  • anonymous
What is the topic that you got these questions from?
gorica
  • gorica
powers and roots
anonymous
  • anonymous
What does the directions for these questions say?
gorica
  • gorica
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.
anonymous
  • anonymous
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.
gorica
  • gorica
sure, if u find solution - post it :)
anonymous
  • anonymous
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.
anonymous
  • anonymous
Put these solution in the equation you are trying to find and solve from there. #1: Answer=2
gorica
  • gorica
but I have to write how did I get it o.O
anonymous
  • anonymous
Since, x^2+y^2+z^2=2 -1<=x<=1 , -1<=y<=1 and -1<=z<=1
anonymous
  • anonymous
From, x+y+z=0 u will get one of them is -1 , the other one is 1 and the other must be 0
anonymous
  • anonymous
So, x^4+y^4+z^4=2
sirm3d
  • sirm3d
\[\large xy+yz+xz=-\frac{ 1 }{ 4 }a^2\]\[\large xyz=-\frac{1}{4}a^3\]using algebra
anonymous
  • anonymous
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
anonymous
  • anonymous
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
anonymous
  • anonymous
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,\((2)^2=x^4+y^4+z^4 + 2(1)\) \(x^4+y^4+z^4 =2\)
gorica
  • gorica
thank you, guys :)
anonymous
  • anonymous
You're welcome :)

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