across
  • across
\[\begin{align} x^1+y^1+z^1&=0\\ x^3+y^3+z^3&=3\\ x^5+y^5+z^5&=15\\\\ x^2+y^2+z^2&=\text{?} \end{align}\]
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
(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?
Mr.Math
  • Mr.Math
\(6\), I think.
anonymous
  • anonymous
Working?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

across
  • across
Mr. Math, how did you get that number?
Mr.Math
  • Mr.Math
As 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
  • apoorvk
is MrM. right across?? cause i seem to be getting '4'??
apoorvk
  • apoorvk
i 'll recheck, i can make a lot of calculating errors. :
Mr.Math
  • Mr.Math
I will recheck too! :)
Mr.Math
  • Mr.Math
There'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
  • across
Mr. Math is right. :) This can be solved using Newton's identities*. Since \(x+y+z=0\), they are the roots of \(t^3+at-b=0\). 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
  • Mr.Math
I don't know Newton's identities, I will read the link you gave. Nice problem by the way :)
apoorvk
  • apoorvk
whoa...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!!!!! :))
anonymous
  • anonymous
For Newton Identities this (http://www.artofproblemsolving.com/Resources/Papers/PolynomialsAK.pdf) is a good reference.

Looking for something else?

Not the answer you are looking for? Search for more explanations.