輪換三元二次方程式

Question

給定變數
x^2+xy+y^2=4

y^2+yz+z^2=9

z^2+xz+x^2=16

求xy+yz+zx=?

Answer 1

let
a = x^2 + xy + y^2 = 4
b= y^2 + yz + z^2 = 9
c = z^2 + xz + x^2 = 16
=> { [ 2*( a * b + b * c + a * c) - a^2 - b^2 - c^2 ] / 3 } ^ ( 1 / 2 )
= xy + yz + xz
湊出來的...自己動手做做看吧
=> { [ 2*( 36 + 144 + 64 ) - 16 - 81 - 256 ] / 3 }^( 1 / 2 )
= 3 * ( 5 )^( 1 / 2 )