Not so efficient!

First, consider the expansion of $(x+y)^n$ , using Binomial Theorem

$(x+y)^n = {n \choose 0}x^n + {n \choose 1}x^{n-1}y + {n \choose 2}x^{n-2}y^2 \ldots {n \choose n}y^n$

Now, we need to just consider the general term of the series, which is $T_{r+1} = {n \choose r} x^{n-r}y^{r}$

Similarly, the general term in the expansion of $(x+3x^{-2})^8$ , would be ${8 \choose r}x^{8-r}(3x^{-2})^r = {8 \choose r}x^{8-3r}3^r$

But, we need to find the the coefficient of $x^2$ . Therefore, $8-3r=2 \Rightarrow r=2$

Hence, the coefficient of $x^2$ is ${8 \choose 2}3^2 = 28\times9 = \boxed{252}$

$(x + 3x^{-2})^{8}$

$= (x + \frac{3}{x^{2}})^{8}$

$= (\frac{x^{3} + 3}{x^{2}})^{8}$

$= \frac{1}{x^{16}} \times (x^{3} + 3)^{8}$

In order to obtain the term involving $x^{2}$ , we have to find the term $x^{18} = (x^{3})^{6}$ in the bracket as it is finally divided by $x^{16}$ .

So, the answer is the coefficient of $(x^{3})^{6}$ in $(x^{3} + 3)^{8}$

As we know, the coefficient of $x^{k}$ in $(x + a)^{n}$ is $_{n - k}^{n}\textrm{C} \times a^{k}$

Hence, in our case we have to find the $6$ th term, whose coefficient is

$_{8 - 6}^{8}\textrm{C} \times 3^{2}$

$= _{2}^{8}\textrm{C} \times 3^{2}$

$= \frac{8!}{2! \times 6!} \times 9$

$= 252$

That's the answer!

Note that we can apply binomial theorem to this problem to solve it. We are attempting to craft $x^2$ by raising $x$ to the $p$ power and $x^{-2}$ to the $q$ power. Note that $p+q=8$ for the sake of binomial theorem. We note that the only solution to this is $p=6$ and $q=2$ because that will result in a net exponent of $2$ . Thus, we apply binomial theorem to find that our answer is $8 \choose {2}$ $* 3^{2} =252.$

Find the $T_{3}$ term = 8 $C_{r}$ x $a^{8-r}$ x $b^{r}$ where r=2 , a=x , b= $3x^{-2}$ and solve...

Alright so first off we need to set up this as a binomial expansion, which is given by

$(a+b)^n = \sum\limits_{i=0}^n$ $n \choose i$ $a^{i}b^{n-i}$

We let $a = x$ , $n = 8$ , and $b = 3x^{-2}$ . Now we have

$\sum\limits_{i=0}^8$ $8 \choose i$ $x^{i}((\frac{3}{x^2})^{8-i}$

Now since we want the coefficient for the $x^2$ term, we need to find out when in the summation runs through a particular $i$ value and out pops an $x^2$ term. If we look at the superscripts of the $x^{i}$ and $(\frac{3}{x^2})^{8-i}$ , we can see that

$(8 - i)(-2) + i = 2$

$i = 6$

And so if you plug in $i = 6$ into the binomial expansion/summation, you will get $\boxed{252}$ as the coefficient.

(x + 3.x^-2)^8 = (x^2 + 6.x^-1 + 9.x^-4 )^4 = (x^4 + 36.x^-2 + 81.x^-8 + 12x + 18.x^-2 + 108.x^-5)^2 = 36x^2 + 18x^2 + ... + 36x^2 + ... + 144x^2 + .... + 18x^2 ---> (36+36+18+18+144)x^2 = (36.3 + 36.4)x^2 = (36.7)x^2 = 252x^2 ---> [252]x^2.. So, the coefficient of the term involving x^2 in the expansion (x + 3.x^-2)^8 is 252. Answer : 252