Suppose x5 occurs in the (r + 1)th term of the expansion (x+3)8
We know that the (r + 1)th term in the expansion of (x+a)n is given by
Tr+1 =

n

‌

Crxn−rar
∴ Tr+1 =

8

‌

C1(x)8−r(3)r ... (i)
On comparing power of x in x5 and Tr+1, we get
8 - r = 5 ⇒ r = 3
Putting r = 3 in (i), we obtain the coefficient of x5 in the expansion of
(x+3)8 =

8

‌

C3(3)3
=

8

‌

C3 . 27 =

8×7×6

3×2

× 27 = 1512