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"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, $$\frac{t^{4012}+37}{2}$$

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  • $\begingroup$ Your question makes no sense. Please edit it. root makes no sense; $t^4 012$ makes no sense. Etc. $\endgroup$ – David G. Stork Jun 15 at 17:14
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    $\begingroup$ By ${\rm root}(4012)(...)$, do you mean $\sqrt[4012]{...} = (...)^{1/4012}$? If that is the case, use the construct \sqrt[4012]{...} and/or (...)^{1/4012} to format them. $\endgroup$ – achille hui Jun 15 at 17:16
  • $\begingroup$ Hint: Calculate $(3\sqrt3-2\sqrt7)^2$. $\endgroup$ – Barry Cipra Jun 15 at 17:23
  • $\begingroup$ Thanks for the hint $\endgroup$ – Ankit Kumar Jun 15 at 17:48
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$t^{4012}=(55+12\sqrt{21})(3\sqrt{3}-2\sqrt{7})^2=(55+12\sqrt{21})(27+28-12\sqrt{21})=(55+12\sqrt{21})(55-12\sqrt{21})=3025-144 \cdot 21=3025-3024=1$

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

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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 $\sqrt[2006]{a}^{4012}=(\sqrt[2006]{a}^{2006})^2= a^2$.

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

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