Question

Find each of the following product:
$(\left(\frac{1}{2}\right)y^2-\left(\frac{1}{3}\right)y)(\left(\frac{1}{2}\right)y^2-\left(\frac{1}{3}\right)y)$

Answer 1

Given that $(\left(\frac{1}{2}\right)y^2-\left(\frac{1}{3}\right)y)(\left(\frac{1}{2}\right)y^2-\left(\frac{1}{3}\right)y)$

But we can write the given expression as

$(\left(\frac{1}{2}\right)y^2-\left(\frac{1}{3}\right)y)\times (\left(\frac{1}{2}\right)y^2-\left(\frac{1}{3}\right)y)={(\left(\frac{1}{2}\right)y^2-\left(\frac{1}{3}\right)y)}^2$

But we have $(a-b{)}^2=a^2-2ab+b^2$

On applying above identity in the given expression we get,

${(\left(\frac{1}{2}\right)y^2-\left(\frac{1}{3}\right)y)}^2={\left(\left(\frac{1}{2}\right)y^2\right)}^2-2\left(\left(\frac{1}{2}\right)y^2\right)\times \left(\left(\frac{1}{3}\right)y\right)+{\left(\left(\frac{1}{3}\right)y\right)}^2$

$\therefore {(\left(\frac{1}{2}\right)y^2-\left(\frac{1}{3}\right)y)}^2=\left(\frac{1}{4}\right)y^4-\left(\frac{1}{3}\right)y^3+\left(\frac{1}{9}\right)y^2$

$\therefore {(\left(\frac{1}{2}\right)y^2-\left(\frac{1}{3}\right)y)}^2=\frac{1}{4}{y}^4-\frac{1}{3}{y}^3+\frac{1}{9}{y}^2$