As we know that any variable raised to the power of zero yields 1, a constant polynomial can be represented as $a{x}^0$, where, $a$ is the constant and $x$ is the variable of the polynomial. Thus the degree of a constant polynomial is 0.

$x+x^{-1}=11$

$({x+x^{-1})}^2={11}^2$  [Squaring both the sides]

$\Rightarrow {\left(x\right)}^2+{\left(x^{-1}\right)}^2+2\left(x\right)\left(x^{-1}\right)=121$

$\Rightarrow x^2+x^{-2}+2=121$   $[\because (x)\left(x^{-1}\right)=x^0=1]$

$\Rightarrow x^2+x^{-2}=121-2=119$

Let
$p\left(x\right)=a{x}^3+b{x}^2+x-6$

By using factor theorem, $(x+2)$ can be a factor of $p\left(x\right)$ only when $p(-2)=0$
$p(-2)=a{(-2)}^3+b{(-2)}^2+(-2)-6=0$       
$\Rightarrow -8a+4b-8=0$                    
$\therefore -2a+b=2...\left(i\right)$

Also when $p\left(x\right)$ is divided by $(x-2)$ the remainder is 4.

$\therefore p\left(2\right)=4$
$\Rightarrow a{\left(2\right)}^3+b{\left(2\right)}^2+2-6=4$
$\Rightarrow 8a+4b+2-6=4$
$\Rightarrow 8a+4b=8$
$\Rightarrow 2a+b=2...\left(ii\right)$

Adding equations (i) and (ii), we get
$(-2a+b)+(2a+b)=2+2$
$\Rightarrow 2b=4\Rightarrow b=2$

Putting $b=2$ in (i) we get

$-2a+2=2$ 
$\Rightarrow -2a=0\Rightarrow a=0$
Hence,
$a=0andb=2$

Since ${(a+b+c)}^2=a^2+b^2+c^2+2(ab+bc+ca)$
$a+b+c=8\text{and}ab+bc+ca=20$
${\left(8\right)}^2=a^2+b^2+c^2+2\times 20$
$64=a^2+b^2+c^2+40$
 $a^2+b^2+c^2=24$
We now use the following identity:
$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-(ab+bc+ca\left)\right)$
$a^3+b^3+c^3-3abc=8\times (24-20)=4\times 8=32$
Thus, $a^3+b^3+c^3-3abc=32$

$x-5=0$ signifies that $(x-5)$ is a polynomial whose value is zero for a certain value of $x$ and this $x$ will be the root of the given equation

$\therefore \text{at x=5 the value of expression x-5 is 0.}$
Hence, the root of the given equation is 5.

Let $x^{\frac{1}{3}}=a,y^{\frac{1}{3}}=b,z^{\frac{1}{3}}=c$
It is given that $a+b+c=0$
$\Rightarrow a+b=-c$
Cubing both sides, we get,
$(a+b{)}^3=(-c{)}^3$ 
$\Rightarrow a^3+b^3+3ab(a+b)=(-c{)}^3$
$\Rightarrow a^3+b^3+3ab(-c)=-c^3$
$\Rightarrow a^3+b^3+c^3=3abc$
Putting back the values of $a,b,c$ in the above equation, we have
$(x^{\frac{1}{3}}{)}^3+(y^{\frac{1}{3}}{)}^3+(z^{\frac{1}{3}}{)}^3=3x^{\frac{1}{3}}{y}^{\frac{1}{3}}{z}^{\frac{1}{3}}$
$\Rightarrow x+y+z=3x^{\frac{1}{3}}{y}^{\frac{1}{3}}{z}^{\frac{1}{3}}$
$\Rightarrow {(x+y+z)}^3=(3x^{\frac{1}{3}}{y}^{\frac{1}{3}}{z}^{\frac{1}{3}}{)}^3$
                           $=27xyz$

Let two numbers be a and b.
The difference of their cubes is $a^3–b^3$ and hence this is the expression we need.
This expression can be factorised as $a^3–b^3=(a-b)(a^2+ab+b^2)$.
Now, consider the given factor, $x^4+x^{2}y+y^2={\left(x^2\right)}^2+x^{2}y+{\left(y\right)}^2$
This is in the form of $a^2+ab+b^2$ where $a=x^2,b=y$, which is similar to the factor given. 

Therefore, the other factor is  of the form $a–b$.   

Hence, other factor is $x^2-y.$

The point where the polynomial cuts the x-axis is the zero of the polynomial. Since the given polynomial cuts the x-axis at 3 points, the number of zeroes is 3.

Since $(x+1)$ is a factor, $x=-1$ will make the given expression zero.

$\therefore (-1{)}^{100}+2\times (-1{)}^{99}+k=0\Rightarrow 1-2+k=0\Rightarrow k=1$