MHB Possible integer values for coefficients of cubic equation with given root

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The discussion centers on finding integer coefficients \( (a, b, c, d) \) for the cubic equation \( ax^3 + bx^2 + cx + d = 0 \) with the root \( x = \sqrt[3]{\sqrt{8}+4} - \sqrt[3]{\sqrt{8}-4} \). Participants explore the implications of this root and its algebraic properties. The conversation highlights the importance of integer solutions and the relationships between the coefficients. The contributors express appreciation for each other's insights and efforts in solving the problem. Overall, the focus remains on determining valid integer values for the coefficients based on the specified root.
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Given $a,\,b,\,c$ and $d$ are all integers such that $x=\sqrt[3]{\sqrt{8}+4}-\sqrt[3]{\sqrt{8}-4}$ is a root to the equation $ax^3+bx^2+cx+d=0$. Find the possible values for $(a,\,b,\,c,\,d)$.
 
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anemone said:
Given $a,\,b,\,c$ and $d$ are all integers such that $x=\sqrt[3]{\sqrt{8}+4}-\sqrt[3]{\sqrt{8}-4}$ is a root to the equation $ax^3+bx^2+cx+d=0$. Find the possible values for $(a,\,b,\,c,\,d)$.

$x=\sqrt[3]{\sqrt{8}+4}-\sqrt[3]{\sqrt{8}-4}$
cube both sides to get
$x^3 = \sqrt{8}+4 - (\sqrt{8}- 4) - 3\sqrt[3]((\sqrt{8}+4)(\sqrt{8}-4))x$
or $x^3=8-3\sqrt[3](8-16)x=8+6x$
or $x^3- 6x -8=0$ so $a = t, b = 0, c = -6t , d= -8t $ where t is any non zero integer
 
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$$\begin{align*}x^3 &= \sqrt{8} + 4 - \sqrt{8} + 4 + 3\left( -\left(\sqrt[3]{\sqrt{8} + 4}\right)^2 \sqrt[3]{\sqrt{8} - 4} + \sqrt[3]{\sqrt{8} + 4} \left(\sqrt[3]{\sqrt{8} - 4}\right)^2 \right) \\
&= 8 + 3\sqrt[3]{\sqrt{8} + 4} \sqrt[3]{\sqrt{8} - 4}\left( -x \right) \\
&= 8 - 3\sqrt[3]{8-16}x \\
&= 8 + 6x \end{align*}$$

Thus

$$a = t,\ b = 0,\ c = -6t,\ d = -8t,\ t \in \mathbb{R} \smallsetminus 0.$$
 
Hi kaliprasad and Theia!

Very well done to the both of you! And thanks for participating!
 
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