A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing
 2 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)
 2 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)

This Question is Open

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

philo1234
 2 years ago
Best ResponseYou've already chosen the best response.0R u there? If so let's look at each problem.

philo1234
 2 years ago
Best ResponseYou've already chosen the best response.0What is the topic that you got these questions from?

philo1234
 2 years ago
Best ResponseYou've already chosen the best response.0What does the directions for these questions say?

gorica
 2 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.

philo1234
 2 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
 2 years ago
Best ResponseYou've already chosen the best response.0sure, if u find solution  post it :)

philo1234
 2 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.

philo1234
 2 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
 2 years ago
Best ResponseYou've already chosen the best response.0but I have to write how did I get it o.O

sauravshakya
 2 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

sauravshakya
 2 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
 2 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

sauravshakya
 2 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

sauravshakya
 2 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

Shadowys
 2 years ago
Best ResponseYou've already chosen the best response.1LOL 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,\((2)^2=x^4+y^4+z^4 + 2(1)\) \(x^4+y^4+z^4 =2\)
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.