 # simple equation uva

http://uva.onlinejudge.org/external/115/11571.html
Can anyone help me with a solution to this problem just not able to get it
Thanks in advance (x+y+z)/xyz = (1/yz)+(1/xz)+(1/xy)
(x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+xz)

i could get to that extent

thanks, but it didnt help me still not got to the solution…not anywhere near it still not got to the solution…not anywhere near it glance links above, that is all i can do for now

hi,
plz leme know if it helps you.

x + y + z = A
xyz = B
x2 + y2 + z2 = C

z^3 - A(Z^2) + {(A^2 - C)/2}z - B = 0 …1 eq

find out using formula for 3 degree equation and will get 3 roots. lets say z1,z2,z3

now equation converted to 2 degree equation :

x + y + z1 = A ==> x + Y = A - z1 = P (say)
xyz1=B ==> xy = B/z1 = Q (say)
x^2 + y^2 + z1^2 = C ==> x^2 y^2 = c - z1^2 = R (say)

so equations are now:

x + Y = P
xy = Q
x^2 y^2 = R

y^2 -Py + Q = 0 …2 eq

values of y can be find out using standered 2 degree equation … let say y1,y2

@parashar13 thanks…but the logic and code of finding roots of a 3rd degree equation is a little hard to remember
the following link is for getting roots of cubic equation
@garakchy thanks 