Question

The number of integral terms in the expansion of ${(7^{1/3}+{11}^{1/9})}^{6561}$ is

Answer 1

$T_{r+1}={}^{6561}{C}_r\left(7^{\frac{1}{3}}{)}^{6561-r}\right({11}^{\frac{1}{9}}{)}^r={}^{6561}{C}_r{7}^{(2187-\frac{r}{3})}{11}^{\left(\frac{r}{9}\right)};$
$r$ must be multiple of $9.$
$r=0,9,18,\cdots ,6561\Rightarrow 6561=0+(k-1)\times 9\Rightarrow k=730$