To find the coefficient of \(x^5\) in the expansion of \(\left(x^2 - \frac{a}{x}\right)^7\), we use the binomial theorem.
The general term in the expansion is given by:
\(T_k = \binom{7}{k} (x^2)^{7-k} \left(-\frac{a}{x}\right)^k\)
We need the term where the power of \(x\) is 5:
\((x^2)^{7-k} \left(-\frac{a}{x}\right)^k = x^{2(7-k)} \cdot (-a)^k \cdot x^{-k} = x^{14-2k-k} = x^{14-3k}\)
Set the exponent equal to 5:
\(14 - 3k = 5\)
Solving for \(k\):
\(14 - 5 = 3k\)
\(9 = 3k\)
\(k = 3\)
Substitute \(k = 3\) into the general term:
\(T_3 = \binom{7}{3} (x^2)^4 \left(-\frac{a}{x}\right)^3\)
\(= \binom{7}{3} x^8 \cdot (-a)^3 \cdot x^{-3}\)
\(= \binom{7}{3} (-a)^3 x^5\)
The coefficient of \(x^5\) is:
\(\binom{7}{3} (-a)^3 = -280\)
Calculate \(\binom{7}{3}\):
\(\binom{7}{3} = 35\)
Thus:
\(35(-a)^3 = -280\)
\((-a)^3 = -8\)
\(a^3 = 8\)
\(a = 2\)
