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

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

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

• Your question makes no sense. Please edit it. root makes no sense; $t^4 012$ makes no sense. Etc. – David G. Stork Jun 15 at 17:14
• By ${\rm root}(4012)(...)$, do you mean $\sqrt{...} = (...)^{1/4012}$? If that is the case, use the construct \sqrt{...} and/or (...)^{1/4012} to format them. – achille hui Jun 15 at 17:16
• Hint: Calculate $(3\sqrt3-2\sqrt7)^2$. – Barry Cipra Jun 15 at 17:23
• Thanks for the hint – Ankit Kumar Jun 15 at 17:48

$$t^{4012}=(55+12\sqrt{21})(3\sqrt{3}-2\sqrt{7})^2=(55+12\sqrt{21})(27+28-12\sqrt{21})=(55+12\sqrt{21})(55-12\sqrt{21})=3025-144 \cdot 21=3025-3024=1$$
Therefore $$\frac{t^{4012}+37}{2}=\frac{1+37}{2}=\frac{38}{2}=19$$
You know I assume that, for any $$a$$ and $$n$$, $$\sqrt[n]{a}^n=\sqrt[n]{a^n}=a$$. So in particular $$\sqrt{a}^{4012}=a$$. Moreover $$4012=2006*2$$, so $$\sqrt{a}^{4012}=(\sqrt{a}^{2006})^2= a^2$$.
$$3\sqrt3-2\sqrt7<0,$$ which says that a real value of $$t$$ does not exist and the needed value does not exist.