Question

The coefficient of $x^3$ in the expansion of $\left(x^2 - \frac{2}{x}\right)^6$ is

• A. $-160$
• B. $-80$
• C. $-40$
• D. $0$
• E. $-10$

Hint:

Equate the exponent of $x$ in the general term to the required power.

Answer 1

The Correct Option is A

Concept: General term of binomial expansion: \[ T_{k+1} = \binom{n}{k} a^{n-k} b^k \]

Step 1: Identify components \[ a = x^2, b = -\frac{2}{x}, n=6 \]

Step 2: General term \[ T_{k+1} = \binom{6}{k} (x^2)^{6-k} \left(-\frac{2}{x}\right)^k \]

Step 3: Simplify powers \[ (x^2)^{6-k} = x^{12-2k}, \left(-\frac{2}{x}\right)^k = (-2)^k x^{-k} \] \[ T_{k+1} = \binom{6}{k} (-2)^k x^{12-3k} \]

Step 4: Find required power \[ 12 - 3k = 3 \Rightarrow k=3 \]

Step 5: Coefficient \[ \binom{6}{3}(-2)^3 = 20 \times (-8) = -160 \] \[ \boxed{-160} \]