1, 2, 3... sequence can be represented using the polynomial $x$.
1, 3, 5... sequence​ can be represented using the polynomial $2x+1$.
2, 4, 6... sequence​ can be represented using the polynomial $2x$.
1, 4, 9... sequence​ can be represented using the polynomial $x^2$.
All of these are polynomials in one variable i.e., x.

$x+\frac{1}{x}=2$ 
Squaring both sides, we get $x^2+\frac{1}{x^2}+2.x.\frac{1}{x}=4$
So, $x^2+\frac{1}{x^2}=4-2=2$
Again Squaring both sides, we get $x^4+\frac{1}{x^4}+2.x^2.\frac{1}{x^2}=4$
$\Rightarrow x^4+\frac{1}{x^4}=4-2=2$
Similarly,  $x^8+\frac{1}{x^8}=x^{32}+\frac{1}{x^{32}}=x^{64}+\frac{1}{x^{64}}=2$

We know that
$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)---\left(1\right)$
According to question, $(x^2+y^2+z^2)=(xy+yz+zx)$.
$\therefore$ RHS will be zero in equation (1)
So,$x^3+y^3+z^3=3xyz$

$p\left(x\right)=a{x}^3-b{x}^2+cx$
If (x - 1) is a factor then x = 1 is the zero of polynomial
p(1) = 0

$p\left(1\right)=a-b+c=0$

It can be written as $a+(-b)+c=0$

We know that if $a+b+c=0,then{a}^3+b^3+c^3=3abc$

So, $a^3+{(-b)}^3+c^3=3a(-b)c=-3abc$

$(x^2-y^2)=(x-y)(x+y)=18$

$(x-y)=3----\left(i\right)$
$(x+y)=\frac{18}{3}=6----\left(ii\right)$

Adding equation (i) and (ii) we get
$2x=9$
$x=\frac{9}{2}$ 
Substituting the value of x in equation (ii) we get
$y=\frac{3}{2}$
$16x^2{y}^2=16\times (\frac{9}{2}{)}^2\times (\frac{3}{2}{)}^2=729$

$x^g$+$\frac{1}{x^a}=x^g+x^{-a}$
Since g = a,
$x^g+\frac{1}{x^a}=x^a+x^{-a}$
For the expression  $x^a+x^{-a}$ to be a polynomial, a and -a must be whole number so that the expression $\frac{1}{x^a}$ = 1 , hence a should be equal to 0 .
Thus, a = 0.
When a=0, the given polynomial will get simplified further as:
$x^g+\frac{1}{x^a}=x^a+x^{-a}$
$\Rightarrow$ $x^g+\frac{1}{x^a}=x^0+x^{-0}$
$\Rightarrow$ $x^g+\frac{1}{x^a}=x^0+x^0$
$\Rightarrow$ $x^g+\frac{1}{x^a}=1+1=2$
'2' is a constant polynomial in one variable with degree zero . Because $2=2x^0$.
So, 'a' must be equal to zero so that given expression is a polynomial in variable.

In this expression, $a^2{x}^2$ and $ax$ are the two terms. Since it is given that 'a' is a constant and 'x' is a variable, we can see that there is only one variable 'x' in the polynomial thus making it a polynomial in one variable.

We have to break up $7x$ in the given expression into two parts in such a way that their sum is $7x$ and the product of their coefficients is $12$ .
$x^2-7x+12$
$=x^2-4x-3x+12$
$=x(x-4)-3(x-4)$
$=(x-3)(x-4)$
Thus $x^2-7x+12$ can be factorised as $(x-3)(x-4)$