Question

multiply the conjugates.
$\left(\frac{1}{3}y - 2\
ight)\left(\frac{1}{3}y + 2\
ight)=$

Explanation:

Step1: Apply difference - of - squares formula

Use $(a - b)(a + b)=a^{2}-b^{2}$, where $a=\frac{1}{3}y$ and $b = 2$.

Step2: Calculate $a^{2}$ and $b^{2}$

$a^{2}=(\frac{1}{3}y)^{2}=\frac{1}{9}y^{2}$, $b^{2}=2^{2}=4$.

Step3: Get the result

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

Answer:

$\frac{1}{9}y^{2}-4$