1. **State the problem:** Find the constant term in the expansion of $$\left(x - \frac{2}{x}\right)^4 \left(x^2 + \frac{2}{x}\right)^3.$$\n\n2. **Recall the binomial expansion formula:** For any integers $n \geq 0$, $$(a+b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k.$$\n\n3. **Expand each binomial separately:**\n- For $$\left(x - \frac{2}{x}\right)^4,$$ the general term is $$\binom{4}{k} x^{4-k} \left(-\frac{2}{x}\right)^k = \binom{4}{k} (-2)^k x^{4-k-k} = \binom{4}{k} (-2)^k x^{4-2k}.$$\n- For $$\left(x^2 + \frac{2}{x}\right)^3,$$ the general term is $$\binom{3}{m} (x^2)^{3-m} \left(\frac{2}{x}\right)^m = \binom{3}{m} 2^m x^{2(3-m)-m} = \binom{3}{m} 2^m x^{6-3m}.$$\n\n4. **Multiply the general terms:**\n$$\binom{4}{k} (-2)^k x^{4-2k} \times \binom{3}{m} 2^m x^{6-3m} = \binom{4}{k} \binom{3}{m} (-2)^k 2^m x^{(4-2k)+(6-3m)} = \binom{4}{k} \binom{3}{m} (-2)^k 2^m x^{10 - 2k - 3m}.$$\n\n5. **Find the constant term:** The power of $x$ must be zero, so solve for $k,m$ in $$10 - 2k - 3m = 0.$$\nRearranged: $$2k + 3m = 10.$$\n\n6. **Find integer solutions $(k,m)$ with $0 \leq k \leq 4$, $0 \leq m \leq 3$:**\n- For $m=0$: $2k=10 \Rightarrow k=5$ (not valid, $k \leq 4$)\n- For $m=1$: $2k + 3 = 10 \Rightarrow 2k=7 \Rightarrow k=3.5$ (not integer)\n- For $m=2$: $2k + 6 = 10 \Rightarrow 2k=4 \Rightarrow k=2$ (valid)\n- For $m=3$: $2k + 9 = 10 \Rightarrow 2k=1 \Rightarrow k=0.5$ (not integer)\n\nOnly valid solution: $(k,m) = (2,2).$\n\n7. **Calculate the coefficient for $(k,m) = (2,2)$:**\n$$\binom{4}{2} \binom{3}{2} (-2)^2 2^2 = 6 \times 3 \times 4 \times 4 = 6 \times 3 \times 16 = 288.$$\n\n**Final answer:** The constant term in the expansion is **288**.