To find the term independent of x, we need the power of x to be zero in the expansion of \(\left( 4x^3 + \frac{1}{2x} \right)^8\).

Consider the general term in the binomial expansion:

\(T_r = \binom{8}{r} (4x^3)^{8-r} \left( \frac{1}{2x} \right)^r\).

Simplifying, we get:

\(T_r = \binom{8}{r} \cdot 4^{8-r} \cdot x^{3(8-r)} \cdot \frac{1}{2^r} \cdot x^{-r}\).

Combine the powers of x:

\(x^{3(8-r) - r} = x^{24 - 4r}\).

We want the power of x to be zero:

\(24 - 4r = 0\).

Solving for r gives:

\(r = 6\).

Substitute \(r = 6\) back into the expression for the term:

\(T_6 = \binom{8}{6} \cdot 4^2 \cdot \frac{1}{2^6}\).

Calculate \(\binom{8}{6} = 28\), \(4^2 = 16\), and \(2^6 = 64\).

Thus, \(T_6 = 28 \cdot 16 \cdot \frac{1}{64} = 28\).

Therefore, the term independent of x is 28.

To find the term independent of x, we use the binomial expansion formula:

\(\left( a + b \right)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\).

Here, \(a = 2x\) and \(b = \frac{1}{x^2}\), and \(n = 6\).

The general term is:

\(T_k = \binom{6}{k} (2x)^{6-k} \left( \frac{1}{x^2} \right)^k\).

Simplifying, we get:

\(T_k = \binom{6}{k} \cdot 2^{6-k} \cdot x^{6-k} \cdot x^{-2k}\).

For the term to be independent of x, the power of x must be zero:

\(6-k - 2k = 0\).

Solving for \(k\), we get:

\(6 - 3k = 0\)

\(3k = 6\)

\(k = 2\).

Substitute \(k = 2\) back into the expression for \(T_k\):

\(T_2 = \binom{6}{2} \cdot 2^{4} \cdot x^{0}\).

\(T_2 = 15 \cdot 16 \cdot 1\).

\(T_2 = 240\).

Thus, the term independent of x is 240.

To find the term independent of x in the expansion of \(\left( x - \frac{1}{x^2} \right)^9\), we use the binomial theorem:

\(\left( x - \frac{1}{x^2} \right)^9 = \sum_{k=0}^{9} \binom{9}{k} x^{9-k} \left( -\frac{1}{x^2} \right)^k\)

The general term is:

\(T_k = \binom{9}{k} x^{9-k} \left( -\frac{1}{x^2} \right)^k = \binom{9}{k} (-1)^k x^{9-k-2k}\)

\(= \binom{9}{k} (-1)^k x^{9-3k}\)

We need the exponent of x to be zero for the term to be independent of x:

\(9 - 3k = 0\)

\(3k = 9\)

\(k = 3\)

Substitute \(k = 3\) into the general term:

\(T_3 = \binom{9}{3} (-1)^3 x^{9-3 \times 3}\)

\(= \binom{9}{3} (-1)^3 x^0\)

\(= \binom{9}{3} (-1)^3\)

\(= 84 \times (-1)\)

\(= -84\)

To find the term independent of x in the expansion of \(\left( x + \frac{3}{x} \right)^4\), we use the binomial theorem:

\(\left( x + \frac{3}{x} \right)^4 = \sum_{r=0}^{4} \binom{4}{r} x^{4-r} \left( \frac{3}{x} \right)^r\)

The general term is:

\(T_r = \binom{4}{r} x^{4-r} \left( \frac{3}{x} \right)^r = \binom{4}{r} x^{4-r} \cdot \frac{3^r}{x^r} = \binom{4}{r} \cdot 3^r \cdot x^{4-2r}\)

We need the exponent of x to be zero for the term to be independent of x:

\(4 - 2r = 0\)

Solving for r gives:

\(4 = 2r\)

\(r = 2\)

Substitute r back into the general term:

\(T_2 = \binom{4}{2} \cdot 3^2 \cdot x^{4-4} = \binom{4}{2} \cdot 9\)

\(\binom{4}{2} = 6\)

\(T_2 = 6 \cdot 9 = 54\)

Thus, the value of the term independent of x is 54.

(a) To expand \((1 + a)^5\), use the binomial theorem:

\((1 + a)^5 = \sum_{k=0}^{5} \binom{5}{k} a^k\).

Calculate the terms up to \(a^3\):

\(\binom{5}{0} a^0 = 1\)

\(\binom{5}{1} a^1 = 5a\)

\(\binom{5}{2} a^2 = 10a^2\)

\(\binom{5}{3} a^3 = 10a^3\)

Thus, \((1 + a)^5 = 1 + 5a + 10a^2 + 10a^3 + \ldots\)

(b) To expand \([1 + (x + x^2)]^5\), substitute \(a = x + x^2\) into the expansion from part (a):

\(1 + 5(x + x^2) + 10(x + x^2)^2 + 10(x + x^2)^3 + \ldots\)

Calculate each term up to \(x^3\):

\(5(x + x^2) = 5x + 5x^2\)

\(10(x + x^2)^2 = 10(x^2 + 2x^3 + \ldots) = 10x^2 + 20x^3 + \ldots\)

\(10(x + x^2)^3 = 10(x^3 + \ldots) = 10x^3 + \ldots\)

Combine terms: \(1 + 5x + (5x^2 + 10x^2) + (20x^3 + 10x^3) + \ldots\)

\(= 1 + 5x + 15x^2 + 30x^3 + \ldots\)