Polynomial Equation Under The Moon Light

A straightforward computation...we will compute $b$ first. $b=\left(\frac{1}{x}+x\right)^3+\left(\frac{1}{x}+x-1\right)^2$ $=\frac{1}{x^3}+\frac{3}{x}+3x+x^3+\frac{1}{x^2}+x^2+1+2-\frac{2}{x}-2x$ $=\frac{1}{x^3}+\frac{1}{x^2}+\frac{1}{x}+3+x+ x^2+x^3=2$ (divide the given equation by $x^3$ ). Now let $y=x+\frac{1}{x}$ . Then $a=y(y+2)(y-1)=y^3+y^2-2y=y^3+(y-1)^2-1=b-1=1$ $a+b=1+2=\boxed{3}$

$x^6+x^5+x^4+x^3+x^2+x+1 = 0 \\ \Rightarrow x^3+x^2+x+1+\dfrac{1}{x} + \dfrac{1}{x^2} + \dfrac{1}{x^3} = 0 \\ \left(x^3+3x+\frac{3}{x} + \frac{1}{x^3} \right) - 3x- \frac{3}{x} + \left(x^2+2+ \frac{1}{x^2} \right) - 2 + \left(x + \frac{1}{x} \right) + 1 = 0 \\ \left(x + \dfrac{1}{x} \right)^3 + \left(x + \dfrac{1}{x} \right)^2 - 2\left(x + \dfrac{1}{x} \right) -1 = 0$

Now we have:

$\left(x + \dfrac{1}{x} \right) \left(x + \dfrac{1}{x} + 2 \right) \left(x + \dfrac{1}{x} -1 \right) \\ = \left(x + \dfrac{1}{x} \right) \left[ \left(x + \dfrac{1}{x} \right)^2 + \left(x + \dfrac{1}{x} \right) -2 \right] \\ = \left(x + \dfrac{1}{x} \right)^3 + \left(x + \dfrac{1}{x} \right)^2 - 2\left(x + \dfrac{1}{x} \right) = 1 \quad \Rightarrow a = 1$

$\left(x + \dfrac{1}{x} \right)^3 + \left(x + \dfrac{1}{x} - 1 \right)^2 \\ = \left(x + \dfrac{1}{x} \right)^3 + \left(x + \dfrac{1}{x} \right)^2 -2 \left(x + \dfrac{1}{x} \right) + 1 = 2 \quad\Rightarrow b = 2$

Therefore, $\quad a+b = 1 + 2 = \boxed{3}$

$x^6 + x^5 + x^4 + x^3 + x^2 +x + 1 = 0 \Rightarrow x^7 - 1 = 0$

So, $x^7 = 1$

Thus, $\frac{x^6}{1} + x + \frac{x^5}{1} + x^2 + \frac{x^4}{1} + x^3 + 1 = 0$

$\Rightarrow \frac{x^6}{x^7} + x + \frac{x^5}{x^7} + x^2 + \frac{x^4}{x^7} + x^3 + 1 = 0$

$\Rightarrow (\frac{1}{x} + x) + (\frac{1}{x^2} + x^2) + (\frac{1}{x^3} + x^3) + 1 =0$

$\Rightarrow (\frac{1}{x} +x) + ((\frac{1}{x}+x)^2 -2) + ((\frac{1}{x}+x)^3 - 3(\frac{1}{x}+x)) + 1 =0$

$\Rightarrow (\frac{1}{x} + x)^3 + (\frac{1}{x} + x)^2 - 2 (\frac{1}{x} + x) - 1 = 0$ [eq. 1]

$\Rightarrow (\frac{1}{x} + x)((\frac{1}{x} + x)^2 + (\frac{1}{x} + x) - 2) - 1 = 0$

$\Rightarrow (\frac{1}{x} + x)(\frac{1}{x} + x + 2)(\frac{1}{x} + x - 1) - 1 = 0$

$\Rightarrow (\frac{1}{x} + x)(\frac{1}{x} + x + 2)(\frac{1}{x} + x - 1) = 1 = a$

Using [eq 1] $\Rightarrow (\frac{1}{x} + x)^3 + (\frac{1}{x} + x)^2 -2(\frac{1}{x} + x) + 1 = 2$

$\Rightarrow (\frac{1}{x} + x)^3 + (\frac{1}{x} +x - 1)^2 = 2 = b$

Therefore, $a+b = 1 + 2 = 3$