"Stretching" Polynomials

Conveniently bashing works: Roots of $x^2-2x-1$ are $1+- \sqrt{2}$ . Taking $(1+\sqrt{2})^6$ into the sextic function works out nicely to $70\sqrt{2}+ 99 + a + a\sqrt{2} + b = 0$ . $a$ must equal -70 to eliminate the irrational part of the equation. Then $99 - 70 + b = 0$ . $b$ must equal -29. Then, |-70 -29| = $\boxed{99}$

From problem above, we can use long division

$\frac{x^6+ax+6}{x^2-2x-1}=(x^4+2x^3+5x^2+12x+29)+ \frac{(a+70)x+(b+29)}{x^2-2x-1}$

It's mean,

${x^6+ax+6}=(x^4+2x^3+5x^2+12x+29){(x^2-2x-1)}+[(a+70)x+(b+29)]$

Since $x^2-2x-1=0$ are two real root, we must have remaind $[(a+70)x+(b+29)]=0.$ $\iff{a+70=0,b+29=0} \implies a=-70,b=-29$

$\mid{a+b} \mid = \boxed{99}$

$x^2-2x-1 = 0 \\ x^2 = 2x+1 \\ x^3 - 2x^2 - x = 0 \\ x^3 - 2(2x+1) - x = 0 \\ x^3 - 5x - 2 = 0$

Repeat this process until the correct degree and form has been reached. It follows that $a=-70$ and $b=-29$ , so $|a+b| = \boxed{99}$ .

Extra: Technically, we haven't shown how this answer guarantees only these two real roots. How can we show or prove that this 6th degree polynomial does in fact only have these two real roots?