Find $x^{10}y^6z^{-21}$ in $(3x^2 -5y^{\frac{1}{2}} + z^{-3})^{24}$ with Binomial theorem. Find $x^{10}y^6z^{-21}$ and $x^{16}y^4z^{-18}$ in $(3x^2 -5y^{\frac{1}{2}} + z^{-3})^{24}$ with Binomial theorem.
Well, I believe I found $x^{10}y^6z^{-21}$ and $x^{16}y^4z^{-18}$ .
But, I'm a bit hesitated about the solution, because I didn't use the fact: $i+j+k = 24$ and I believe I have to use it.  What do you guys think? Here's my solution:

 A: It is also convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ in an algebraic expression.

Instead of using the multinomial version we could also successively apply the binomial theorem. We obtain
\begin{align*}
[x^{10}&y^6z^{-21}](3x^2-5y^{\frac{1}{2}}+z^{-3})^{24}\tag{1}\\
&=[x^{10}y^6z^{-21}]\sum_{k=0}^{24}\binom{24}{k}3^kx^{2k}(-5y^{\frac{1}{2}}+z^{-3})^{24-k}\tag{2}\\
&=3^5\binom{24}{5}[y^6z^{-21}](-5y^{\frac{1}{2}}+z^{-3})^{19}\\
&=3^5\binom{24}{5}[y^6z^{-21}]\sum_{k=0}^{19}\binom{19}{k}(-5)^ky^{\frac{1}{2}k}z^{-3(19-k)}\tag{3}\\
&=3^5(-5)^{12}\binom{24}{5}\binom{19}{12}\\
\end{align*}

Comment:


*

*In (2) we select $k=5$ in order to obtain the coefficient of $[x^{10}]$.

*In (3) we select $k=12$ in order to obtain the coefficient of $[y^{6}]$ as well as $[z^{-21}]$.

Note: If we consider a summand $$[x^{10}y^6z^{-21}]a_{ijk}x^{2i}y^{\frac{1}{2}j}z^{-3k}$$ of the trinomial (1) with coefficient $a_{ijk}$, the following equations have to be fulfilled in order to find a possible solution
  \begin{align*}
i+j+k&=24\\
2i&=10\\
\frac{1}{2}j&=6\\
-3k&=-21
\end{align*}
The last three equations give $i=5,j=12$ and $k=7$ fulfilling $i+j+k=24$.



We start (as we could always do) the second example with a plausibility check regarding the exponents
\begin{align*}
[x^{16}&y^4z^{-18}](3x^2-5y^{\frac{1}{2}}+z^{-3})^{24}
\end{align*}
We get the following equations
\begin{align*}
i+j+k&=24\\
2i&=16\\
\frac{1}{2}j&=4\\
-3k&=-18
\end{align*}
The last three equations give $i=8,j=8$ and $k=6$ not fulfilling $i+j+k=24$.
We conclude
  \begin{align*}
[x^{16}&y^4z^{-18}](3x^2-5y^{\frac{1}{2}}+z^{-3})^{24}=0
\end{align*}

