$(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})abc=\frac{bc+ac+ab}{abc}\ast <!--\; \ast \; -->\frac{abc}{1}=bc+ac+ab$

$5x^2(\frac{1}{x^2}-<!--\; -\; -->\frac{1}{x}+3)=5x^2\ast <!--\; \ast \; -->\frac{1-<!--\; -\; -->x+3x^2}{x^2}=5(1-<!--\; -\; -->x+3x^2)=5-<!--\; -\; -->5x+15x^2$

$(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})\ast <!--\; \ast \; -->\frac{ab}{c}=\frac{a^{2}c+b^{2}c+c^{2}b}{abc}\ast <!--\; \ast \; -->\frac{ab}{c}=\frac{c(a^2+b^2+cb)}{c\ast <!--\; \ast \; -->c}=\frac{a^2+b^2+cb}{c}$

$3x^3(\frac{2}{x^2}+\frac{1}{y}+\frac{4}{x})=3x^3\ast <!--\; \ast \; -->\frac{2y+x^2+4xy}{x^{2}y}=\frac{3x(2y+x^2+4xy)}{y}$