Find the coefficient of $x^3$ in the expansion of ${(3x+4)}^6$

${(3x+4)}^6={(3x^1+\frac{4}{x^0})}^6$

$a=3,p=1,b=4,q=0,n=6$

$k=3\therefore \text{coefficient of }{x}^3$

$\therefore r=\frac{6-3}{1}=3$

Coefficient of $x^3$ is ${}^6{\text{C}}_3{\left(3\right)}^{6-3}\times {\left(4\right)}^3$

$={}^6{\text{C}}_3{\left(3\right)}^3{\left(4\right)}^3$

$={}^6{\text{C}}_3\times 27\times 64$

$=20\times 27\times 64$

$=34,560$