Algebra

G'day mate, 

"y" in this case is "a", I accidentally used "a" throughout, sorry!!!!!

\(\begin{bmatrix}\frac{1}{x}+\frac{1}{a}=\frac{1}{3}\\ xa+x+a=4\end{bmatrix}\)

\(\mathrm{Plug\:the\:solutions\:}x=-\frac{3a}{3-a}\mathrm{\:into\:}xa+x+a=4\)

We get a = \(a=\frac{1}{2}+i\frac{\sqrt{11}}{2} .and. a=\frac{1}{2}-i\frac{\sqrt{11}}{2}\)

\(\mathrm{Plug\:the\:solutions\:}a=\frac{1}{2}-i\frac{\sqrt{11}}{2},\:a=\frac{1}{2}+i\frac{\sqrt{11}}{2}\mathrm{\:into\:}\frac{1}{x}+\frac{1}{a}=\frac{1}{3}\)

That gives \(x=\frac{1}{2}+\frac{\sqrt{11}}{2}i and x=\frac{1}{2}-\frac{\sqrt{11}}{2}i\)

So now we have our values for (x, y): \(\begin{pmatrix}x=\frac{1}{2}+\frac{\sqrt{11}}{2}i,\:&y=\frac{1}{2}-i\frac{\sqrt{11}}{2}\\ x=\frac{1}{2}-\frac{\sqrt{11}}{2}i,\:&y=\frac{1}{2}+i\frac{\sqrt{11}}{2}\end{pmatrix}\)

Plugging that in: \(x^2\cdot y + x\cdot y^2\):

\(\left(\frac{1}{2}+\frac{\sqrt{11}}{2}i\right)^2\left(\frac{1}{2}-\frac{\sqrt{11}}{2}i\right)+\left(\frac{1}{2}+\frac{\sqrt{11}}{2}i\right)\left(\frac{1}{2}-\frac{\sqrt{11}}{2}i\right)^2=3\)

\(\left(\frac{1}{2}-\frac{\sqrt{11}}{2}i\right)^2\left(\frac{1}{2}+\frac{\sqrt{11}}{2}i\right)+\left(\frac{1}{2}-\frac{\sqrt{11}}{2}i\right)\left(\frac{1}{2}+\frac{\sqrt{11}}{2}i\right)^2=3\)

So the answer is \(3.\)

-Vinculum 

1/x + 1/y  =  1/3     ⇒   (x + y) / xy  =   1/3   ⇒  (x + y)  = (1/3)xy  ⇒ 3(x + y) = xy

And

xy + ( x + y) = 4

3xy + 3(x + y) = 12

3xy + xy = 12

4xy = 12

xy  = 3

Since (x +y) / xy  =  1/3   this implies that  (x + y)   =1

So

x^2y  + y^2x   =

xy ( x + y)   =

3 ( 1)   = 3