If x and y are positive integers such that  xy+x+y = 71,  x²y +xy² =880, then x ²+y² is equal to

Solution:

$\because \mathrm{xy}+\mathrm{x}+\mathrm{y}=71\Rightarrow \mathrm{xy}+(\mathrm{x}+\mathrm{y})=71$

and  ${\mathrm{x}}^2\mathrm{y}+{\mathrm{xy}}^2=880\Rightarrow \mathrm{xy}(\mathrm{x}+\mathrm{y})=880$

$\Rightarrow \mathrm{xy} \text{and }(\mathrm{x}+\mathrm{y})$ are the roots of the quadratic equation. 

$\begin{array}{rl}{\mathrm{t}}^2-71\mathrm{t}+880& =0\\ \Rightarrow (\mathrm{t}-55)(\mathrm{t}-16)& =0\\ \Rightarrow \mathrm{t}& =55,16\\ \mathrm{x}+\mathrm{y}& =16\text{and}\mathrm{xy}=55\end{array}$

So,${\mathrm{x}}^2+{\mathrm{y}}^2=(\mathrm{x}+\mathrm{y}{)}^2-2\mathrm{xy}=(16{)}^2-110=146$