A community for students.
Here's the question you clicked on:
 0 viewing
across
 3 years ago
\[\begin{align}
x^1+y^1+z^1&=0\\
x^3+y^3+z^3&=3\\
x^5+y^5+z^5&=15\\\\
x^2+y^2+z^2&=\text{?}
\end{align}\]
across
 3 years ago
\[\begin{align} x^1+y^1+z^1&=0\\ x^3+y^3+z^3&=3\\ x^5+y^5+z^5&=15\\\\ x^2+y^2+z^2&=\text{?} \end{align}\]

This Question is Closed

AsianDuck
 3 years ago
Best ResponseYou've already chosen the best response.0(x+y+z)(x+y+z) = 0 > x^2 + y^2 + z^2 + 2xy + 2xz + 2yz = 0 > x^2 + y^2 + z^2 = 2(xy + xz + yz) Is this even in the right direction?

across
 3 years ago
Best ResponseYou've already chosen the best response.3Mr. Math, how did you get that number?

Mr.Math
 3 years ago
Best ResponseYou've already chosen the best response.3As Asian started \(x^2+y^2+z^2=2(xy+xz+yz)\) \((a)\) We have from the first and the second equation that \(x^3+y^3(x+y)^3=3\) \( \implies x^2y+xy^2=1\), and by symmetry we also have \(x^2z+zx^2=1\) and \(y^2z+yz^2=1\). We can also write \[(x^2+y^2+z^2)(x^3+y^3+z^3)=3(x^2+y^2+z^2)\] \[(x^5+y^5+z^5)+xy(xy^2+yx^2)+xz(xz^2+zx^2)+yz(yz^2+zy^2)\] \(\implies 3(x^2+y^2+z^2)=15(xy+xz+yz)\) \((b)\) By solving (a) and (b), it's easy to find \(x^2+y^2+z^2=6\).

apoorvk
 3 years ago
Best ResponseYou've already chosen the best response.0is MrM. right across?? cause i seem to be getting '4'??

apoorvk
 3 years ago
Best ResponseYou've already chosen the best response.0i 'll recheck, i can make a lot of calculating errors. :

Mr.Math
 3 years ago
Best ResponseYou've already chosen the best response.3There's a missing equal sign there. It should be \[(x^2+y^2+z^2)(x^3+y^3+z^3)=3(x^2+y^2+z^2)\] \(=(x^5+y^5+z^5)+xy(xy^2+yx^2)+xz(xz^2+zx^2)+yz(yz^2+zy^2)\).

across
 3 years ago
Best ResponseYou've already chosen the best response.3Mr. Math is right. :) This can be solved using Newton's identities*. Since \(x+y+z=0\), they are the roots of \(t^3+atb=0\). Newton's identities say that\[x^3+y^3+z^3=3b\]\[x^4+y^4+z^4=2a^2\]and\[x^2+y^2+z^2=2a\]This implies that \(b=1\) and \(2a^2=15\). It is then possible to solve for \(a\) and find the value of \(x^2+y^2+z^2=2a\). Note that \(x^3+y^3+z^3=3b\) was not necessary. Mr. Math's approach is a pedestrian way of achieving this, however. * http://en.wikipedia.org/wiki/Newton%27s_identities

Mr.Math
 3 years ago
Best ResponseYou've already chosen the best response.3I don't know Newton's identities, I will read the link you gave. Nice problem by the way :)

apoorvk
 3 years ago
Best ResponseYou've already chosen the best response.0whoa...i did a very stupid mistake towards the end :/ and now realise that the answer should have been somewhat symmetric (as in a multiple of 3 or some way), 4 would be pretty odd. Nice problem @across, had no clue about "newton's identities" will read that. and @Mr.Math , thanks for doing it before me, and posting the solution!!!!! :))

FoolForMath
 3 years ago
Best ResponseYou've already chosen the best response.0For Newton Identities this (http://www.artofproblemsolving.com/Resources/Papers/PolynomialsAK.pdf) is a good reference.
Ask your own question
Sign UpFind 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.