Answer: \(4096-384x+15x^2\); term independent of \(x\) is \(-8177\).

(a) Use the binomial expansion of

\(\left(4-\frac{x}{16}\right)^6.\)

The first term is

\(4^6=4096.\)

The second term is

\(6(4^5)\left(-\frac{x}{16}\right) =6(1024)\left(-\frac{x}{16}\right) =-384x.\)

The third term is

\({6\choose2}(4^4)\left(\frac{x}{16}\right)^2 =15(256)\frac{x^2}{256} =15x^2.\)

So the first three terms are

\(4096-384x+15x^2.\)

(b) Also

\(\left(x-\frac1x\right)^2=x^2-2+\frac1{x^2}.\)

The constant term comes from two products:

\(4096(-2)\quad\text{and}\quad 15x^2\cdot\frac1{x^2}.\)

Therefore the term independent of \(x\) is

\(-8192+15=-8177.\)