(a) Express \(f(x)\) in partial fractions:

Assume \(\frac{17x^2 - 7x + 16}{(2 + 3x^2)(2 - x)} = \frac{Ax + B}{2 + 3x^2} + \frac{C}{2 - x}\).

Multiply through by the denominator: \(17x^2 - 7x + 16 = (Ax + B)(2 - x) + C(2 + 3x^2)\).

Expand and equate coefficients to solve for \(A, B,\) and \(C\):

\(A = -2, B = 3, C = 5\).

Thus, \(f(x) = \frac{-2x + 3}{2 + 3x^2} + \frac{5}{2 - x}\).

(b) Expand \(f(x)\) in ascending powers of \(x\):

Use the binomial expansion for \((2 - x)^{-1}\) and \((2 + 3x^2)^{-1}\):

\((2 - x)^{-1} = \frac{1}{2} + \frac{x}{4} + \frac{x^2}{8} + \frac{x^3}{16}\).

\((2 + 3x^2)^{-1} = \frac{1}{2} - \frac{3x^2}{4} + \frac{9x^4}{8}\).

Substitute back into the partial fractions and expand:

\(\frac{-2x + 3}{2 + 3x^2} = \left(-2x + 3\right)\left(\frac{1}{2} - \frac{3x^2}{4}\right)\).

\(\frac{5}{2 - x} = 5\left(\frac{1}{2} + \frac{x}{4} + \frac{x^2}{8} + \frac{x^3}{16}\right)\).

Combine and simplify to get:

\(4 + \frac{1}{4}x - \frac{13}{8}x^2 + \frac{29}{16}x^3\).

(c) The expansion is valid for \(|x| < \sqrt[3]{\frac{2}{3}}\).

(i) Express \(f(x)\) in partial fractions:

Assume \(\frac{16 - 17x}{(2 + x)(3 - x)^2} = \frac{A}{2 + x} + \frac{B}{3 - x} + \frac{C}{(3 - x)^2}\).

Multiply through by the denominator \((2 + x)(3 - x)^2\) to clear the fractions:

\(16 - 17x = A(3 - x)^2 + B(2 + x)(3 - x) + C(2 + x)\).

Expand and equate coefficients to solve for \(A, B,\) and \(C\):

\(A = 2, \ B = -2, \ C = -1\).

(ii) Expand \(f(x)\) in ascending powers of \(x\):

Use the partial fraction decomposition:

\(\frac{2}{2 + x} = 1 - \frac{x}{2} + \frac{x^2}{4} + \cdots\)

\(\frac{-2}{3 - x} = -\frac{2}{3} - \frac{2x}{9} - \frac{2x^2}{27} + \cdots\)

\(\frac{-1}{(3 - x)^2} = -\frac{1}{9} - \frac{2x}{27} - \frac{3x^2}{81} + \cdots\)

Add these expansions up to the \(x^2\) term:

\(f(x) = \left(1 - \frac{x}{2} + \frac{x^2}{4}\right) + \left(-\frac{2}{3} - \frac{2x}{9} - \frac{2x^2}{27}\right) + \left(-\frac{1}{9} - \frac{2x}{27} - \frac{3x^2}{81}\right)\).

Combine like terms:

\(f(x) = \frac{8}{9} - \frac{43}{54}x + \frac{7}{108}x^2\).

(i) Express \(f(x)\) in partial fractions:

Assume \(\frac{12 + 12x - 4x^2}{(2+x)(3-2x)} = \frac{A}{2+x} + \frac{B}{3-2x}\).

Multiply through by \((2+x)(3-2x)\) to clear the denominators:

\(12 + 12x - 4x^2 = A(3-2x) + B(2+x)\).

Expand and collect terms:

\(12 + 12x - 4x^2 = (3A + 2B) + (-2A + B)x\).

Equate coefficients to solve for \(A\) and \(B\):

\(3A + 2B = 12\) and \(-2A + B = 12\).

Solving these equations gives \(A = 2\), \(B = -4\), and \(C = 6\).

Thus, \(f(x) = \frac{2}{2+x} + \frac{-4}{3-2x}\).

(ii) Expansion in ascending powers of \(x\):

Expand \(\frac{1}{2+x}\) using the binomial series:

\(\frac{1}{2+x} = \frac{1}{2}(1 - \frac{x}{2} + (\frac{x}{2})^2 + \ldots) = \frac{1}{2} - \frac{x}{4} + \frac{x^2}{8} + \ldots\).

Expand \(\frac{1}{3-2x}\) using the binomial series:

\(\frac{1}{3-2x} = \frac{1}{3}(1 + \frac{2x}{3} + (\frac{2x}{3})^2 + \ldots) = \frac{1}{3} + \frac{2x}{9} + \frac{4x^2}{27} + \ldots\).

Substitute back into the partial fractions:

\(f(x) = 2(\frac{1}{2} - \frac{x}{4} + \frac{x^2}{8}) - 4(\frac{1}{3} + \frac{2x}{9} + \frac{4x^2}{27})\).

Simplify to obtain:

\(f(x) = 2 + \frac{7}{3}x + \frac{7}{18}x^2\).

(i) Express \(f(x)\) in partial fractions:

Assume \(\frac{7x^2 - 15x + 8}{(1 - 2x)(2 - x)^2} = \frac{A}{1 - 2x} + \frac{B}{2 - x} + \frac{C}{(2 - x)^2}\).

Equating coefficients, solve:

\(7 = A + 2B\)

\(-15 = -4A - 5B - 2C\)

\(8 = 4A + 2B + C\)

Solving gives \(A = 1\), \(B = 3\), \(C = -2\).

(ii) Expand \(f(x)\) in ascending powers of \(x\):

Expand each term:

\(\frac{1}{1 - 2x} = 1 + 2x + 4x^2\)

\(\frac{1}{2 - x} = \frac{1}{2}(1 + \frac{x}{2} + \frac{x^2}{4})\)

\(\frac{1}{(2 - x)^2} = \frac{1}{4}(1 + x + \frac{3x^2}{2})\)

Combine terms:

\(f(x) = \frac{1}{1 - 2x} + 3 \cdot \frac{1}{2 - x} - 2 \cdot \frac{1}{(2 - x)^2}\)

\(= (1 + 2x + 4x^2) + 3(\frac{1}{2} + \frac{x}{4} + \frac{x^2}{8}) - 2(\frac{1}{4} + \frac{x}{2} + \frac{3x^2}{8})\)

Combine and simplify:

\(= \frac{9}{4} + x + 4x^2\)

(i) To express \(f(x)\) in the form \(\frac{A}{3-x} + \frac{Bx+C}{2+x^2}\), we equate:

\(\frac{x - 4x^2}{(3-x)(2+x^2)} = \frac{A}{3-x} + \frac{Bx+C}{2+x^2}\)

Multiply through by \((3-x)(2+x^2)\) to clear the denominators:

\(x - 4x^2 = A(2+x^2) + (Bx+C)(3-x)\)

Expand and collect like terms:

\(x - 4x^2 = 2A + Ax^2 + 3Bx + Cx - Bx^2 - Cx\)

\(x - 4x^2 = (A-B)x^2 + (3B+C)x + 2A\)

Equating coefficients, we get:

\(A - B = -4\)

\(3B + C = 1\)

\(2A = 0\) so \(A = 0\)

Substitute \(A = 0\) into \(A - B = -4\):

\(-B = -4\) so \(B = 4\)

Substitute \(B = 4\) into \(3B + C = 1\):

\(12 + C = 1\) so \(C = -11\)

Thus, \(A = 0\), \(B = 4\), \(C = -11\).

(ii) Expand \(\frac{1}{3-x}\) and \(\frac{4x-11}{2+x^2}\) using binomial series:

\(\frac{1}{3-x} = \frac{1}{3} \left(1 + \frac{x}{3} + \frac{x^2}{9} + \frac{x^3}{27} \right)\)

\(\frac{4x-11}{2+x^2} = \frac{4x-11}{2} \left(1 - \frac{x^2}{2} + \frac{x^4}{4} \right)\)

Combine and simplify terms up to \(x^3\):

\(\frac{1}{6}x - \frac{11}{18}x^2 - \frac{31}{108}x^3\)