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if $x+\frac{1} {x} =1$ then $x^2-x+1=0$

$(x+1)(x^2-x+1)$ is a factor of $x(1+x^3 +x^6 +x^9 +x^{18} +x^{21}+x^{24} +x^{30} +x^{33} +x^{39} )$

$(x)(x+1)(x^2-x+1)(1 + x^6 + x^{18} + x^{24} - x^{27} + 2 x^{30} - x^{33} + x^{36})$

but $x^2-x+1=0$

So ( $x)(x+1)(0)(1 + x^6 + x^{18} + x^{24} - x^{27} + 2 x^{30} - x^{33} + x^{36})= \boxed 0$

the roots of the equation $x+ \dfrac{1}{x}=1$ are $\quad -\omega \quad and\quad -\omega^2\quad$ where $\quad \omega,\omega^2 \quad$ are cubic roots of unity $$ note that $\begin{cases} \omega^{x}= \omega , x=1\mod 3\\ \omega^{x}= \omega^2 , x=2\mod 3\\ \omega^{x}= 1 , x=0\mod 3\\ \end{cases}$ now , insert the value $-\omega+{( -\omega )}^{ 4 }+{ (-\omega) }^{ 7 }+{ (-\omega) }^{ 10 }+{( -\omega) }^{ 19 }+{( -\omega) }^{ 22 }+{( -\omega) }^{ 25 }+{( -\omega) }^{ 31 }+{ (-\omega) }^{ 34 }+{ (-\omega) }^{ 40 }$ from the cases, we get that the expression is $-\omega+\omega-\omega+\omega-\omega+\omega-\omega-\omega+\omega+\omega$ the positive cancel out the negatives and we are left with $\boxed{0}$

$x+\frac{1}{x}=-1$

or, $x^2+x+1=0$

or, $(x+1)(x^2+x+1)=0$ or, $x^3+1=0$ or, $x^3=-1$ .

Now, $1729+x+x^{4}+x^{7}+x^{10}+x^{19}+x^{22}+x^{25}+x^{31}+x^{34}+x^{40}$

$=1729+x-x+x-x+x-x+x+x-x-x= \boxed{1729}$