No $x$ Since Birth

Rewrite the expression as $\frac{(x^{3}+x^{2} + 1)^{10}}{x^{10}}$ . Now, we need to find the coefficient of $x^{10}$ in the numerator for the the term that doesn't contain $x$ . Only two possibilities, either take $x^{3}$ and $x^{2}$ both twice and rest $1$ , or take $x^{2}$ five times and rest $1s$ .

Hence the answer is $\overset { 10 }{ \underset { 2 }{ C } } .\overset { 8 }{ \underset { 2 }{ C } } + \overset { 10 }{ \underset { 5 }{ C } } = \boxed{1512}$

Any term in the expanded expression can be written as $C(x^2)^a(x)^b(x^{-1})^c$ where $a+b+c=10$ and $a,b,c$ are non-negative integers. Since we wish to find a term independent on $x$ , we get $2a+b-c=0$ . Adding the two equations yields $3a+2b=10$ which have non-negative integer solutions $(a,b,c)=(0,5,5),(2,2,6)$ . By multinomial theorem, its coefficient (which is the term itself) equals to $\frac{10!}{0!5!5!}+\frac{10!}{2!2!6!}=1512$ .

$\displaystyle \left( x^2+x+\frac {1}{x} \right)^{10} = \left( x(x+1)+\frac {1}{x} \right)^{10} = \sum _{n=0} ^{10} {\begin{pmatrix} 10 \\ n \end{pmatrix} \left( x(x+1) \right) ^{10-n}x^{-n}}$

$\displaystyle = \sum _{n=0} ^{10} {\begin{pmatrix} 10 \\ n \end{pmatrix} x^{10-2n} (x+1) ^{10-n}}$

There are only two $n$ that have $x^0$ terms:

$\begin{cases} n = 5 & \Rightarrow \begin{pmatrix} 10 \\ 5 \end{pmatrix} x^{0} (x+1) ^{5} & \Rightarrow a_{01} = \begin{pmatrix} 10 \\ 5 \end{pmatrix} = 252 \\ n = 6 & \Rightarrow \begin{pmatrix} 10 \\ 6 \end{pmatrix} x^{-2} (x+1) ^{4} & \Rightarrow a_{02} = \begin{pmatrix} 10 \\ 6 \end{pmatrix} \begin{pmatrix} 4 \\ 2 \end{pmatrix} = 1260 \end{cases}$

Therefore, $a_0 = a_{01} + a_{02} = \boxed{1512}$

{ X^2+X+1/X}^10 = { (1/X+X) + X^2 }^10...
. . X^2 term will be X to even powers only. To neutralist this, the terms in the parenthesis should be negative even. We expanding for even powers only ..............
........................ .nCm means combination

Expanding 1st term

(1/X+X)^10 ......only middle term will not have X. So it is 10C5 = 10!/ (5! * 5!) = 252.

Expanding 3rd term
10C2{ (1/X+X)^8 * (X^2)^2 } = 10C2{ (1/X+X)^8 * X^4 }

So the term that give X^(-4) in (X+1/X)^8
X^(-8) + ....... +8C2{ X^(-6) * X^2 +........= ........ +8C2X^(- 4) +...............

Expanded 3rd term = 10C2{ ....+ 8C2X^(- 4) * X^4 + ...................... }
= 10C2 * 8C2 = 1260.
Total = 252 + 1260 =1512