Factors of $a^{2} + \frac{1}{4} + a$ will be

When a polynomial $f(x)$ is divided by $x-3$ and $x+6,$ the respective remainders are $7$ and $22.$ What is the remainder when $f(x)$ is divided by $(x-3)(x+6)?$

Let $f(x) = a_{0}x^{n} + a_{1}x^{n-1} + a_{2}x^{n-2} + \ldots + a_{n-1}x + a_{n}$,where $a_{0}, a_{1}, a_{2}, \ldots, a_{n}$ are constants. If $f(x)$ is divided by $ax - b$,then the remainder is:

If $x+\frac{1}{x}=3,$ then the value of $x^{6}+\frac{1}{x^{6}}$ is

Resolve into factors: $81 x^{2} y^{2}+108 x y z+36 z^{2}$.

If $a+b+c=0,$ then the value of $\left(\frac{a^{2}}{b c}+\frac{b^{2}}{c a}+\frac{c^{2}}{a b}\right)$ is: