Question 2: Do the calculation: \( \frac{1}{2}{x^2}{y^2}\left( {2x + y} \right)\left( {2x – y} \right) \)

We have:

\(\begin{array}{l} \frac{1}{2}{x^2}{y^2}\left( {2x + y} \right)\left( {2x – y} \right) \\ = \frac{1}{2}{x^2}{y^2}(2x.2x + 2x.( – y) + y.2x + y.( – y))\\ = \frac{ 1}{2}{x^2}{y^2}\left( {4{x^2} – 2xy + 2xy – {y^2}} \right)\\ = \frac{1}{2} {x^2}{y^2}\left( {4{x^2} – {y^2}} \right) = \frac{1}{2}{x^2}{y^2}. 4{x^2} + \frac{1}{2}{x^2}{y^2}.\left( { – {y^2}} \right) = 2{x^4}{y^ 2} – \frac{1}{2}{x^2}{y^4} \end{array}\)