I found the following exercise in a textbook:
Exercise 11.2. Find a polynomial with rational coefficients that has $$\frac{1}{3}-\sqrt{2 + \sqrt[5]{1-\sqrt[3]{7}}}$$ as a root.
(Source: Integers, Polynomials, and Rings, by Ronald Irving, p. 178.)
(In what follows, I'll use $A$ to refer to the real number mentioned in the exercise.)
I don't see how one can use any of the material preceding this exercise to solve this problem. Is there a simple algorithm I'm missing?
FWIW, according to Mathematica, the minimal polynomial $f(x)$ in $\mathbb{Z}[x]$ (rather than $\mathbb{Q}[x]$) for $A$ is $$f(x) = 205891132094649 x^{30}\\ -2058911320946490 x^{29}\\ +3774670755068565 x^{28}\\ +26689591197454500 x^{27}\\ +103288717934148915 x^{26}\\ +120743710577284158 x^{25}\\ +979596962250571665 x^{24}\\ +25113634383890520 x^{23}\\ +5534307029830645755 x^{22}\\ +2990868225888744630 x^{21}\\ +21801155360211123894 x^{20}\\ +17387732153625824880 x^{19}\\ +64771042155589328805 x^{18}\\ +57930026664623685390 x^{17}\\ +151839596608592644350 x^{16}\\ +130602662006312722752 x^{15}\\ +287146843258295642445 x^{14}\\ +205985143814386139730 x^{13}\\ +438236071775524498185 x^{12}\\ +219265354586979477900 x^{11}\\ +528283318408667131986 x^{10}\\ +131814318593663712900 x^9\\ +480947808117639761190 x^8\\ +1840154837168719440 x^7\\ +305643459116482257195 x^6\\ +76077403177127438538 x^5\\ +115600935241965908070 x^4\\ +59228312720004507600 x^3\\ +14798700839480423580 x^2\\ +16440663358610960040 x\\ +3233167181917838873$$
It's hard for me to imagine that the textbook's author intended students to compute a solution like that by hand. Is there a (not necessarily minimal) polynomial $g(x)$ in $\mathbb{Q}[x]$, with $g(A) = 0$, that is easier to compute than $f(x)$ above?
$$\begin{array}{rcl} x &=& \displaystyle \frac{1}{3}-\sqrt{2 + \sqrt[5]{1-\sqrt[3]{7}}} \\ \displaystyle x - \frac13 &=& \displaystyle -\sqrt{2 + \sqrt[5]{1-\sqrt[3]{7}}} \\ \displaystyle \left( x - \frac13 \right)^2 &=& \displaystyle 2 + \sqrt[5]{1-\sqrt[3]{7}} \\ \displaystyle \left( x - \frac13 \right)^2 - 2 &=& \displaystyle \sqrt[5]{1-\sqrt[3]{7}} \\ \displaystyle \left( \left( x - \frac13 \right)^2 - 2 \right)^5 &=& \displaystyle 1-\sqrt[3]{7} \\ \displaystyle \left( \left( x - \frac13 \right)^2 - 2 \right)^5 - 1 &=& \displaystyle -\sqrt[3]{7} \\ \displaystyle \left( \left( \left( x - \frac13 \right)^2 - 2 \right)^5 - 1 \right)^3 &=& \displaystyle -7 \\ \displaystyle \left( \left( \left( x - \frac13 \right)^2 - 2 \right)^5 - 1 \right)^3 + 7 &=& 0 \\ \end{array}$$