Evaluate the expression

subtracting both equations..........
(x-y)^{3}=(-20)
(x-y)=(-20)^{\frac{1}{3}}
(x-y)^{2}=(20)^{\frac{2}{3}}
x^{2}+y^{2}-2xy=(20)^{\frac{2}{3}}
let xy be k₁,and x²+y²=k₂
expanding first equation.....
-(20)^1/3.(k₁ + k₂) -3k₁k₂=-20..(1)
k₁ -2k₂=20^2/3
putting k₁=20^2/3+2k₂ in (1)
we get x² +y²=1/2(20^2/3-20^1/3)

squaring both sides and adding .we get x² +y² =10

that's correct(a pink to xyz!).

But there's an easier solution. So this thread's still open

(x+iy)^3 = (x^3-3xy^2)+i(3x^2y-y^3) = 10 - 30i \\ \\ \Rightarrow x^2+y^2 = |x+iy|^2 = |10-30i|^{\frac{2}{3}} = 10

Nice soln Prophet sir.....I applied Geometry to this.....

Let Angle ACB be \theta so I have \frac{x}{y}=tan\theta Putting this in the above equations and dividing them, I have tan3\theta=\frac{-1}{3}.....
Now in triangle ECA,
\frac{\sqrt{x^2+y^2}}{sin3\theta}=\frac{4y}{sin2\theta}
Also I know \frac{y}{\sqrt{x^2+y^2}}=cos\theta
From which I have sin^2\theta=\frac{5}{8}\Rightarrow 3x^2=5y^2
x^2+y^2=\frac{8y^2}{3}
Putting x²=5y²3 in 2nd eqn, we can get y, and can get \frac{8y^2}{3} as well......Done.