Find the term independent of $x$ in the expansion of $(x^{2}+4/x)^{6}$

Question

I figured out that the question can be solved by finding r and then using the formula of $Tr+1$. But the problem is I get stuck in the middle

Answer 1

Hint:

Use the binomial formula: $$\Bigl(x^2+\frac4x\Bigr)^6=\sum_{k=0}^6\binom 6kx^{2k}\frac{4^{6-k}}{x^{6-k}}=\sum_{k=0}^6\binom 6k 4^{6-k}x^{3k-6}.$$

Answer 2

\begin{align*} \left( x^2 + \frac{4}{x} \right)^6 &= \left( \left( x^2 + \frac{4}{x} \right)^2 \right)^3 \\ &= \left( x^4 + 8x + \frac{16}{x^2} \right)^3 \text{.} \end{align*} The degrees are $4$, $1$, and $-2$. Taking these in triples, the only combinations giving $0$ are $4-2-2$ and $1+1-2$. So the constant terms are $$ 3 \cdot (x^4)\left(\frac{16}{x^2}\right)\left(\frac{16}{x^2}\right) = 768 $$ and $$ 3 \cdot (8x)(8x)\left(\frac{16}{x^2}\right) = 3072 \text{.}$$

So the desired coefficient is $768 + 3072 = 3840$.

Answer 3

Note: $$\left( x^2 + \frac{4}{x} \right)^6=\frac{(x^3+4)^6}{x^6}=\frac{(4+x^3)^6}{x^6}=\frac{4^6+6\cdot 4^5\cdot x^3+15\cdot 4^4\cdot (x^3)^2+A(x)}{x^6}=\\ B(x)+15\cdot 4^4=B(x)+3840.$$

Answer 4

$\dfrac{(x^3+4)^6}{x^6}=$

$(\dfrac{1}{x^6}) \displaystyle \sum_{k=0}^{6} \binom{6}{k} (x^3)^{6-k}4^k;$

We need the term $k=4$ in the sum(why?).

Term independent of $x:$

$\displaystyle \binom{6}{4}4^4.$