Question

Find the product of $(\frac{1}{2}m-\frac{1}{3})(\frac{1}{2}m+\frac{1}{3})(\frac{1}{4}{m}^2+\frac{1}{9})$

Answer 1

Using the formula $(a+b)(a-b)$ = $a^2-b^2$
So $(\frac{1}{2}m-\frac{1}{3})(\frac{1}{2}m+\frac{1}{3})$ = $(\frac{1}{2}m{)}^2-(\frac{1}{3}{)}^2$
$=$ $(\frac{1}{4}{m}^2-\frac{1}{9})$
Now we have $(\frac{1}{4}{m}^2-\frac{1}{9})$$(\frac{1}{4}{m}^2+\frac{1}{9})$ $=(\frac{1}{4}{m}^2{)}^2-(\frac{1}{9}{)}^2$
So $=$ $\frac{1}{16}{m}^4-\frac{1}{81}$
The answer is $\frac{1}{16}{m}^4-\frac{1}{81}$