Find $a^{4} +2a^{3}-\frac{2}{a}$

Question

Given that $a^3+2a^2+a+1=0$, Find $a^{4} +2a^{3}-\frac{2}{a}$

I tried resolving the cubuc equation..but to no avail. I got this question from one of my friend and have been trying to solve this question since 3 hrs but have not been able to solve it until now. My friend says that the expression whose value is to be found is 1, but i am unable to reach that conclusion. I also tried plotting on Desmos to get the roots of the given cubic equation. The graph showed that this cubic equation has only one real root and that is approximately equal to $-1.755$. I substituted this value in the expression-in-question but the value of that expression turns out to be something that us not equal to 1.

P.S. Assume that $a$ is real. Please help me solve this problem. Please!

Answer 1

Just for your curiosity !

As said, there is a problem somewhere but it is doable.

The real solution of $$a^3+2a^2+a+1=0$$ is given by $$a=-\frac{4}{3} \cosh ^2\left(\frac{1}{6} \cosh ^{-1}\left(\frac{25}{2}\right)\right)$$ which gives the nice result $$\frac{128}{81} \cosh ^6\left(\frac{1}{6} \cosh ^{-1}\left(\frac{25}{2}\right)\right) \left(\cosh \left(\frac{1}{3} \cosh ^{-1}\left(\frac{25}{2}\right)\right)-2\right)+\frac{3}{2} \text{sech}^2\left(\frac{1}{6} \cosh ^{-1}\left(\frac{25}{2}\right)\right)$$

Answer 2

( Just for fun as well ! )

My friend says that the expression whose value is to be found is 1

It has been established already that the value of the expression is not $1$, so there must be a typo somewhere in the given expressions. Following is a guess about what typos would render $1$ as the correct answer.

Consider for example the related problem.

Given that $a^3 + 2 a^2 + p a + q = 0$ find $p, q \in \mathbb R$ such that $a^{4} +2a^{3}-\cfrac{2}{a} = 1$

Leaving the calculations to WA, the resultant of the two polynomials turns out to factor nicely.

$$ \mathtt{Resultant}\,[\, a^3 + 2 a^2 + p a + q, a^5 + 2 a^4 - 2 - a, a \,] \\ \; = \; -(p - q - 1) (2 p - q) (p + q + 3) \big( (p-1)^2 + (q-2)^2 \big) $$

It follows that the solution set to the posed problem is the family of values:

$$ \{ \; (p, q) \in \mathbb R^2 \;\mid\; p - q = 1 \;\lor\; p + q = -3 \;\lor\; p = 2 q \;\lor\; (p, q) = (1, 2) \} $$

For example, either of the following conditions cause $a^{4} +2a^{3}-\cfrac{2}{a}$ to evaluate to $1$.

$$ \\ a^3 + 2 a^2 + a + 2 = 0 \\ a^3 + 2 a^2 + 2 a + 1 = 0 \\ a^3 + 2 a^2 - a - 2 = 0 \\ a^3 + 2 a^2 - 2 a - 1 = 0 \\ a^3 + 2 a^2 + 2021 a + 2020 = 0 $$

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[ EDIT ]   Tackling the question from the other end, $a^{4} +2a^{3}-\cfrac{2}{a} = 1$ iff $a \in \{ -2, \pm1, \pm i\}$, so any perturbation of the original polynomial $a^3 + 2 a^2 + a + 1$ that makes one of $\{ -2, \pm1, \pm i\}$ to become a root will satisfy the problem. For example, in the case of $a^3 + 2 a^2 + p a + q$ the condition $p - q = 1$ is equivalent to having $-1$ as a root.