Expand the following:
$\left(\frac{1}{x}+\frac{y}{3}\right)^{3}$
$\because \quad(x+y)^{3}=x^{3}+y^{3}+3 x y(x+y)$
$\left(\frac{1}{x}+\frac{y}{3}\right)^{3}=\left(\frac{1}{x}\right)^{3}+\left(\frac{y}{3}\right)^{3}+3 \times \frac{1}{x} \times \frac{y}{3}\left(\frac{1}{x}+\frac{y}{3}\right)$
$=\frac{1}{x^{3}}+\frac{y^{3}}{27}+\frac{y}{x}\left(\frac{1}{x}+\frac{y}{3}\right)$
$=\frac{1}{x^{3}}+\frac{y^{3}}{27}+\frac{y}{x^{2}}+\frac{y^{3}}{3 x}=\frac{1}{x^{3}}+\frac{y}{x^{2}}+\frac{y^{2}}{3 x}+\frac{y^{3}}{27}$
Write the degree of each of the following polynomials
$x^{50}-1$
Factorise $: x^{3}-125$
Factorise
$x^{2}+\frac{y^{2}}{4}+\frac{z^{2}}{16}+x y+\frac{y z}{4}+\frac{z x}{2}$
If $\frac{x}{y}+\frac{y}{x}=-1(x, y \neq 0),$ the value of $x^{3}-y^{3}$ is
Find the quotient and the remainder when $x^{3}+x^{2}-10 x+8$ is divided by
$x+3$