Factorise:

$10\left(3 x+\frac{1}{x}\right)^{2}-\left(3 x+\frac{1}{x}\right)-3=10\left(3 x+\frac{1}{x}\right)^{2}-6\left(3 x+\frac{1}{x}\right)+5\left(3 x+\frac{1}{x}\right)-3$

$=\left[2\left(3 x+\frac{1}{x}\right)\right]\left[5\left(3 x+\frac{1}{x}\right)-3\right]+1\left[5\left(3 x+\frac{1}{x}\right)-3\right]$

$=\left[5\left(3 x+\frac{1}{x}\right)-3\right]\left[2\left(3 x+\frac{1}{x}\right)+1\right]$

$=\left(15 x+\frac{5}{x}-3\right)\left(6 x+\frac{2}{x}+1\right)$

Hence, factorisation of $10\left(3 x+\frac{1}{x}\right)^{2}-\left(3 x+\frac{1}{x}\right)-3$ is $\left(15 x+\frac{5}{x}-3\right)\left(6 x+\frac{2}{x}+1\right)$.