If x + y = √3 and x - y = √2, then the value of 8xy (x2 + y2) is

From the given data

We know that x + y = √3 and x -­ y = √2

By adding both the equations, we get

⇒ 2x = √3 + √2

⇒ x = (√3 + √2)/2

Substitute x = (√3 + √2)/2 in x + y = √3

$(\Rightarrow \;y\; = \;2\sqrt 3 - \frac{{\sqrt 3 + \sqrt 2 }}{2}\; = \;\frac{{\sqrt 3 \;--\;\sqrt 2 }}{2})$

Substitute x = (√3 + √2)/2 and y = (√3 - √2)/2 in 8xy (x2 + y2)

$(\begin{array}{l} \Rightarrow \;8xy\left( {{x^2} + {y^2}} \right)\; = \;8 \times \frac{{\sqrt 3 + \sqrt 2 }}{2} \times \frac{{\sqrt 3 - \sqrt 2 }}{2}\;\left[ {{{\left\{ {\frac{{\sqrt 3 + \sqrt 2 }}{2}} \right\}}^2} + {{\left\{ {\frac{{\sqrt 3 - \sqrt 2 }}{2}} \right\}}^2}} \right]\\ = \;8 \times \frac{{3 - 2}}{4} \times \frac{{\left[ {2\left( {3 + 2} \right)} \right]}}{4} \end{array})$

= 5