To find the coefficient of \(\frac{1}{x^3}\) in the expansion of \(\left( x - \frac{2}{x} \right)^7\), we use the binomial theorem:
\(\left( x - \frac{2}{x} \right)^7 = \sum_{k=0}^{7} \binom{7}{k} x^{7-k} \left( -\frac{2}{x} \right)^k\)
We need the term where the power of \(x\) is \(-3\):
\(x^{7-k} \left( -\frac{2}{x} \right)^k = x^{7-k} \cdot (-2)^k \cdot x^{-k} = (-2)^k x^{7-2k}\)
Set the exponent of \(x\) to \(-3\):
\(7 - 2k = -3\)
\(7 + 3 = 2k\)
\(10 = 2k\)
\(k = 5\)
Substitute \(k = 5\) into the binomial coefficient and expression:
\(\binom{7}{5} x^{7-5} \left( -\frac{2}{x} \right)^5 = \binom{7}{5} x^2 (-2)^5 x^{-5}\)
\(= \binom{7}{5} x^{2-5} (-32)\)
\(= \binom{7}{5} x^{-3} (-32)\)
\(= 21 \cdot (-32) x^{-3}\)
\(= -672 x^{-3}\)
Thus, the coefficient of \(\frac{1}{x^3}\) is \(-672\).
