Suppose $x^5$ 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 $T_{r+1}$ = $\,{}^{n}C_r x^{n-r} a^r$ ∴ $T_{r+1}$ = $\,{}^{8}C_1 (x)^{8-r} (3)^r$ ... (i) On comparing power of x in $x^5$ and $T_{r+1}$, we get 8 - r = 5 ⇒ r = 3 Putting r = 3 in (i), we obtain the coefficient of $x^5$ in the expansion of $(x+3)^8$ = $\,{}^{8}C_3(3)^3$ = $\,{}^{8}C_3$ . 27 = $\frac{8 \times 7 \times 6}{3 \times 2}$ × 27 = 1512