Prove that $\sqrt[3]{\sec\frac{2\pi }{7}} + \sqrt[3]{\sec\frac{4\pi }{7}} + \sqrt[3]{\sec\frac{8\pi }{7}} = \sqrt[3]{8-6\sqrt[3]{7}}$

Question

Prove that:

$$\sqrt[3]{\sec\frac{2\pi }{7}} + \sqrt[3]{\sec\frac{4\pi }{7}} + \sqrt[3]{\sec\frac{8\pi }{7}} = \sqrt[3]{8-6\sqrt[3]{7}}$$

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Answer 1

$$ \sec\frac{2\pi}{7},\qquad \sec\frac{4\pi}{7},\qquad \sec\frac{8\pi}{7} $$ are the three zeros of the polynomial $z^3+4z^2-4z-8$ .

The cube-roots of those zeros are the zeros of the polynomial $y^9+4y^6-4y^3-8$ . Using as a hint the right-hand side above, we can factor this polynomial to get a polynomial of degree 6 with no real roots, and a polynomial of degree 3 with three real roots $$ y^3+(6\cdot 7^{1/3} - 8)^{1/3}y^2+\left( \frac{(6\cdot 7^{1/3}-8)^{2/3}(3-4\cdot 7^{1/3}-3\cdot 7^{2/3})}{25} \right)y-2 $$ and therefore the sum of its zeros is, indeed, $$ -(6\cdot 7^{1/3}-8)^{1/3} $$