Squared Radicals

Reorder, we get:

$(\sqrt{60}-3\sqrt{7})(\sqrt{60}+3\sqrt{7})(\sqrt{60}-3\sqrt{7})(\sqrt{60}-3\sqrt{7})$

$((\sqrt{60}-3\sqrt{7})(\sqrt{60}+3\sqrt{7}))^2$

The expression above is a conjugate binomial to square.

$\Rightarrow (\sqrt{60}-3\sqrt{7})(\sqrt{60}+3\sqrt{7})=(\sqrt{60})^2-(3\sqrt{7})^2=(60-63)=-3$

$\therefore ((\sqrt{60}-3\sqrt{7})(\sqrt{60}+3\sqrt{7}))^2=(-3)^2=\boxed{9}$

We now that, if $a, b \in \mathbb{R}\\ a^2b^2=(ab)^2$ \ $(\sqrt{60}+3\sqrt{7})^2 (\sqrt{60}-3\sqrt{7})^2=((\sqrt{60}+3\sqrt{7})(\sqrt{60}-3\sqrt{7}))^2$ \ $((\sqrt{60})^2-(3\sqrt{7})^2)^2=(60-63)^2$ $\Longrightarrow (-3)^2=9$

(a+b)^2 (a-b)^2=a^4+b^4-2ab So (√60)^4+(3√7)^4- 2(√60)^2(3√7)^2. =9

(a+b)^2 (a-b)^2

=a^4+b^4-2ab

so, (√60)^4+(3√7)^4- 2(√60)^2(3√7)^2. =9

((√60)²-3²(√7)²)² = (60-9*7)² = (-3)² = 9

i solve the problem using (a+b) 7 ( a-b) square formula

(60-63)^2=9