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You are given only this information:
x + y + z = 4
x2 + y2 + z2 = 6
xyz = 3
You are required to determine a numerical value for:
x3 + y3 + z3
No fancy stuff, just algebraic manipulation... I couldn't figure it out so anyone who can point me in the right direction, please relieve me of this.
that is so easy!!!
(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)=16
x^2+y^2+z^2=6
=> xy+yz+xz=5
now
(xy+yz+zx)*(x+y+z)=x^2y+xyz+zx^2+xy^2
+y^2z+zxy+xyz+yz^2+z^2x =
x^2y+zx^2+xy^2+y^2z+yz^2+z^2x+3*xyz=20
A=x^2y+zx^2+xy^2+y^2z+yz^2+z^2x=11
A=11
now
(x^2+y^2+z^2)(x+y+z)=
x^3+y^3+z^3 +A =24
and
x^3+y^3+z^3 =13
SEE YA
:o :o :o :o :o
I might have made a stupid mistake somewhere in here, but I can't find one.
Anyways, this should help you along:
http://www.vbforums.com/attachment.p...postid=1146584
how did u do these images, NotLKH?
Yeah thanks a lot, as soon as I read your first lines I realized I'd missed the obvious way of doing it, hehe.
Now for part 2:
Given the drawing, find X:
I had a 5 minute stab at this and couldn't be arsed to try any harder...
its easy to do,hehe :D :D
what are these lines? medians? if its random lines i doubt it can be solved.
They are in fact, random. You are given EXACTLY what I am. The question gives that exact diagram and says "Find the area x." I think it can be done but I don't have the time to try and work it out for myself... also I can't be arsed ;)
It looks to me a bit trig intensive, but definetely do-able.
Too bad I don't have the time.
Here's an angle map of the problem I'd use if I did have time:
http://www.vbforums.com/attachment.p...postid=1147119
is this an equalateral, scalene or isocolese triangle? or do you not have that information?
You only get EXACTLY what I posted: that drawing and the quote that goes with it. You would have to assume that it's scaline I suppose...
I just spent 5 minutes thinking and worked this out....
So now I need the area A
solution is in this thread: http://www.vbforums.com/showthread.p...hreadid=194972