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Calculate the value of the radical: $$\left[\sqrt[{\sqrt[\sqrt3]{\sqrt3}}]{(3 \sqrt 3)^{ \sqrt[\sqrt3]{\sqrt3^{\sqrt3+1}}}}\right]^{\sqrt 3}$$

What is the tip to solve this problem? I think $(3\sqrt{3})^{9}$ is the answer

Attemp: I started thinking about replacing $ \sqrt {3} $ for $y$, but I can't develop much

Lambert macuse
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1 Answers1

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Just do it. But do it slowly and patiently.

$\left[\sqrt[\color{green}{{\sqrt[\sqrt3]{\sqrt3}}}]{(3 \sqrt 3)^{ \color{red}{ \sqrt[\sqrt3]{\sqrt3^{\sqrt3+1}}}}}\right]^{\color{purple}{\sqrt 3}}=$

$(3\sqrt 3)^{\frac {\color{purple}{\sqrt 3}\color{red}{ \sqrt[\sqrt3]{\sqrt3^{\sqrt3+1}}}}{\color{green}{{\sqrt[\sqrt3]{\sqrt3}}}}}=$

$(3\sqrt 3)^{(\sqrt 3)^{\color{purple}{1}-\color{green}{\sqrt 3}}\color{red}{\sqrt{3}^{\frac{\sqrt 3+1}{\sqrt 3}}}}=$

$(3\sqrt 3)^{\sqrt{3}^{1-\sqrt 3+ \frac {\sqrt 3+1}{\sqrt 3}}}=$

$(3\sqrt 3)^{\sqrt{3}^{1-\sqrt 3+ 1+\sqrt 3}}=$

$(3\sqrt 3)^{\sqrt 3^2}=$

$(3\sqrt 3)^3 =$

$(3^13^{\frac 12})^3=$

$(3^{1+\frac 12})^3 =$

$(3^{\frac 32})^3=$

$3^{\frac 92}=$

$3^{4\frac 12}=$

$3^4\sqrt 3=$

$81\sqrt 3$

fleablood
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  • Why $1 - \sqrt{3}$ in the third stage?? – Lambert macuse Apr 23 '21 at 11:06
  • The purple $1$ comes from the $\color{purple}{\sqrt 3}=\color{purple}{\sqrt 3^1}$ in the numerator. The green $\sqrt 3$ comes from the $\color{green}{\sqrt 3^{\sqrt 3}}$ in the denominator. Combining them we get $(\sqrt 3)^{\color{purple}1 - \color{green}{\sqrt 3}}$. – fleablood Apr 23 '21 at 16:18