Question
Expand: $\left(\frac{3}{x}-\frac{4 x}{3}\right)^4$
✓
Answer
$\left(\frac{3}{x}-\frac{4 x}{3}\right)^4= \left(4 C_0\left(\frac{3}{x}\right) 4 \cdot\left(\frac{4 x}{3}\right) 0\right)-4 C_1\left(\frac{3}{x}\right) 3 \cdot\left(\frac{4 x}{3}\right) 1+4 C_0\left(\frac{3}{x}\right) 2 \cdot\left(\frac{4 x}{3}\right) 2-4 C_3\left(\frac{3}{x}\right) 1 \cdot\left(\frac{4 x}{3}\right) 3$
$\quad+4 C_4\left(\frac{3}{x}\right) 0 \cdot\left(\frac{4 x}{3}\right) 4$
$= 1 \cdot \frac{81}{x^4} \cdot 1-4 \cdot \frac{27}{x^3} \cdot \frac{4 x}{3}+\frac{4 x 3}{2 x 1} \cdot \frac{9}{x^2} \cdot \frac{16 x^2}{9}-4 \cdot \frac{3}{x} \cdot \frac{64 x^3}{27}+1 \cdot 1 \cdot \frac{256 x^4}{81}$
$= \frac{81}{x^4}-\frac{144}{x^2}+96-\frac{256 x^2}{9}+\frac{256 x^4}{81}$
$\quad+4 C_4\left(\frac{3}{x}\right) 0 \cdot\left(\frac{4 x}{3}\right) 4$
$= 1 \cdot \frac{81}{x^4} \cdot 1-4 \cdot \frac{27}{x^3} \cdot \frac{4 x}{3}+\frac{4 x 3}{2 x 1} \cdot \frac{9}{x^2} \cdot \frac{16 x^2}{9}-4 \cdot \frac{3}{x} \cdot \frac{64 x^3}{27}+1 \cdot 1 \cdot \frac{256 x^4}{81}$
$= \frac{81}{x^4}-\frac{144}{x^2}+96-\frac{256 x^2}{9}+\frac{256 x^4}{81}$
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