Question

Enter 1 if it is true else 0.
$(\sqrt{2}+3p+3\sqrt{2r}{)}^2-(\sqrt{2}-3p-3\sqrt{2r}{)}^2$ $=12(\sqrt{2}p+2\sqrt{r})$

Answer 1

${(\sqrt{2}+3p+3\sqrt{2r})}^2-{(\sqrt{2}-3p-3\sqrt{2r})}^2$

Using $a^2-b^2=(a-b)(a+b)$ we have

$=(\sqrt{2}+3p+3\sqrt{2r}-\sqrt{2}+3p+3\sqrt{2r})(\sqrt{2}+3p+3\sqrt{2r}+\sqrt{2}-3p-3\sqrt{2r})$

$=(6p+6\sqrt{2r})\left(2\sqrt{2}\right)$

$=12\sqrt{2}(p+\sqrt{2r})$

$=12p\sqrt{2}+24p\sqrt{r}$

$=12(\sqrt{2}p+2\sqrt{r})$