+9674 The complexity of this problem is not the same level as a homework problem, so I will work it out in full details.
Using trinomial theorem (which is just a generalization of binomial theorem), we have
\(\left(2z - \dfrac1{\sqrt z} + z^3\right)^9 = \displaystyle \sum_{\substack{a + b + c = 9\\a,b,c\text{ nonnegative integers}}} \binom{9}{a,b,c} (2z)^a \left(-\dfrac1{\sqrt z}\right)^b (z^3)^c \)
where \(\displaystyle\binom{9}{a,b,c}\) denotes the trinomial coefficient \(\displaystyle\binom{n}{a,b,c} = \dfrac{n!}{a! b! c!}\).
So basically, we first have to find nonnegative integers a, b with a + b <= 9 where a - b/2 + 3(9 - a - b) = 0.
If we move b/2 to the other side of the equation, we get b/2 = a + 3(9 - a - b). Since the right-hand side is an integer, the left-hand side, i.e., b/2, must also be an integer. So we can brute-force and try b = 0, 2, 4, 6, 8 and see which one works. Trying one by one gives a = 10, b = 2 or a = 3, b = 6 as the only solutions with b in the range 0 <= b <= 9. Since a + b <= 9, we can reject the first solution. Doing so, we have \(\begin{cases}a = 3\\b = 6\end{cases}\) as the only nonnegative integer solution to the equation.
Now, the constant term is \(\displaystyle \binom{9}{3,6,0} (2z)^3 \left(-\dfrac1{\sqrt z}\right)^6(z^3)^0 = \binom{9}3 2^3 = 672\).
+118690 (2z - 1/sqrt(z) + z^3)^9.
\((2z - \frac{1}{\sqrt z}+ z^3)^9\)
Is this your intended question?
+9674 The complexity of this problem is not the same level as a homework problem, so I will work it out in full details.
Using trinomial theorem (which is just a generalization of binomial theorem), we have
\(\left(2z - \dfrac1{\sqrt z} + z^3\right)^9 = \displaystyle \sum_{\substack{a + b + c = 9\\a,b,c\text{ nonnegative integers}}} \binom{9}{a,b,c} (2z)^a \left(-\dfrac1{\sqrt z}\right)^b (z^3)^c \)
where \(\displaystyle\binom{9}{a,b,c}\) denotes the trinomial coefficient \(\displaystyle\binom{n}{a,b,c} = \dfrac{n!}{a! b! c!}\).
So basically, we first have to find nonnegative integers a, b with a + b <= 9 where a - b/2 + 3(9 - a - b) = 0.
If we move b/2 to the other side of the equation, we get b/2 = a + 3(9 - a - b). Since the right-hand side is an integer, the left-hand side, i.e., b/2, must also be an integer. So we can brute-force and try b = 0, 2, 4, 6, 8 and see which one works. Trying one by one gives a = 10, b = 2 or a = 3, b = 6 as the only solutions with b in the range 0 <= b <= 9. Since a + b <= 9, we can reject the first solution. Doing so, we have \(\begin{cases}a = 3\\b = 6\end{cases}\) as the only nonnegative integer solution to the equation.
Now, the constant term is \(\displaystyle \binom{9}{3,6,0} (2z)^3 \left(-\dfrac1{\sqrt z}\right)^6(z^3)^0 = \binom{9}3 2^3 = 672\).
!
