The number of irrational terms in the expansion of $${\left( {\root 8 \of 5 + \root 6 \of 2 } \right)^{100}}$$   is

Solution :
$${t_{r + 1}} = {\,^{100}}{C_r}{\left( {\root 8 \of 5 } \right)^{100 - r}} \cdot {\left( {\root 6 \of 2 } \right)^r}.$$
As 2 and 5 are coprime, $${t_{r + 1}}$$ will be rational if $$100 - r$$   is a multiple of 8 and $$r$$ is a multiple of 6. Also $$0 \leqslant r \leqslant 100.$$
$$\eqalign{ & \therefore \,\,r = 0,6,12,.....,96 \cr & \therefore \,\,100 - r = 4,10,16,.....,100\,\,\,\,\,\,.....\left( 1 \right) \cr} $$
But $$100 - r$$   is to be a multiple of 8.
So, $$100 - r = 0,8,16,24,.....,96\,\,\,\,\,\,\,\,.....\left( 2 \right)$$
The common terms in (1) and (2) are 16, 40, 64 and 88.
∴ $$r = 84, 60, 36, 12$$     give rational terms.
∴ the number of irrational terms = $$101 - 4 = 97.$$