Find the value of $\frac{t^{4012}+37}{2}$

Question

"Want a hint for this question as I am not getting it how to start." $$\sqrt[4012]{55+12\sqrt {21}}×\sqrt[2006]{3\sqrt 3 - 2\sqrt 7} = t$$

Then find the value of, $$\frac{t^{4012}+37}{2}$$

Answer 1

$t^{4012}=(55+12\sqrt{21})(3\sqrt{3}-2\sqrt{7})^2=(55+12\sqrt{21})(27+28-12\sqrt{21})=(55+12\sqrt{21})(55-12\sqrt{21})=3025-144 \cdot 21=3025-3024=1$

Therefore $\frac{t^{4012}+37}{2}=\frac{1+37}{2}=\frac{38}{2}=19$

Answer 2

You know I assume that, for any $a$ and $n$, $\sqrt[n]{a}^n=\sqrt[n]{a^n}=a$. So in particular $\sqrt[4012]{a}^{4012}=a$. Moreover $4012=2006*2$, so $\sqrt[2006]{a}^{4012}=(\sqrt[2006]{a}^{2006})^2= a^2$.

Answer 3

$$3\sqrt3-2\sqrt7<0,$$ which says that a real value of $t$ does not exist and the needed value does not exist.