2025 Edexcel IAL Further Mathematics (YFM01) 模擬試題連答案詳解

(a) We can rewrite the given division as multiplication:
\[z_1 = z_2 \left(\frac{1}{2} + \frac{5}{2}\mathrm{i}\right)\]
Substituting the expressions for \(z_1\) and \(z_2\):
\[p + 4\mathrm{i} = (3 - q\mathrm{i})\left(\frac{1}{2} + \frac{5}{2}\mathrm{i}\right)\]
\[p + 4\mathrm{i} = \frac{3}{2} + \frac{15}{2}\mathrm{i} - \frac{q}{2}\mathrm{i} - \frac{5q}{2}\mathrm{i}^2\]
Since \(\mathrm{i}^2 = -1\):
\[p + 4\mathrm{i} = \left(\frac{3 + 5q}{2}\right) + \left(\frac{15 - q}{2}\right)\mathrm{i}\]
Equating the imaginary parts of both sides:
\[4 = \frac{15 - q}{2} \implies 8 = 15 - q \implies q = 7\]

(b) Equating the real parts of both sides:
\[p = \frac{3 + 5q}{2}\]
Substituting \(q = 7\):
\[p = \frac{3 + 5(7)}{2} = \frac{38}{2} = 19\]

(c) Since \(z_1 = 19 + 4\mathrm{i}\), its modulus is:
\[|z_1| = \sqrt{19^2 + 4^2} = \sqrt{361 + 16} = \sqrt{377}\]

(a)
M1: Writes down the equation as \(z_1 = z_2(0.5 + 2.5\mathrm{i})\) or multiplies numerator and denominator of \(\frac{p+4\mathrm{i}}{3-q\mathrm{i}}\) by \(3+q\mathrm{i}\) to equate to \(\frac{1}{2} + \frac{5}{2}\mathrm{i}\).
A1: Correctly expands to find the imaginary part of the product, yielding \(4 = \frac{15-q}{2}\) or equivalent.
A1*: Fully correct argument showing that \(q = 7\).

(b)
M1: Sets up a correct equation for the real part \(p = \frac{3+5q}{2}\) or equivalent, and substitutes their value of \(q\).
A1: Obtains \(p = 19\).

(c)
M1: Correctly applies the modulus formula \(|z_1| = \sqrt{a^2 + b^2}\) on their \(z_1\).
A1: Obtains the exact value \(\sqrt{377}\) (or equivalent exact surd).

From the quadratic equation \(2x^2 - 5x + 4 = 0\), we have:
\[\alpha + \beta = \frac{5}{2}\]
\[\alpha\beta = 2\]

(a)
\[\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\]
\[\alpha^2 + \beta^2 = \left(\frac{5}{2}\right)^2 - 2(2) = \frac{25}{4} - 4 = \frac{9}{4}\]

(b)
Using the identity \(\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)\):
\[\alpha^3 + \beta^3 = \left(\frac{5}{2}\right)^3 - 3(2)\left(\frac{5}{2}\right) = \frac{125}{8} - 15 = \frac{5}{8}\]

(c)
The new equation has roots \(\frac{1}{\alpha^2}\) and \(\frac{1}{\beta^2}\).
Sum of roots:
\[S = \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{\alpha^2\beta^2} = \frac{\frac{9}{4}}{(\alpha\beta)^2} = \frac{\frac{9}{4}}{4} = \frac{9}{16}\]
Product of roots:
\[P = \frac{1}{\alpha^2} \times \frac{1}{\beta^2} = \frac{1}{(\alpha\beta)^2} = \frac{1}{4}\]
Using the equation template \(x^2 - Sx + P = 0\):
\[x^2 - \frac{9}{16}x + \frac{1}{4} = 0\]
Multiplying by 16 to get integer coefficients:
\[16x^2 - 9x + 4 = 0\]

(a)
B1: States or uses both \(\alpha + \beta = \frac{5}{2}\) and \(\alpha\beta = 2\).
M1: Applies the identity \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\) with their values.
A1: Obtains \(\frac{9}{4}\) (or equivalent fraction/decimal).

(b)
M1: Uses a correct identity for \(\alpha^3 + \beta^3\) in terms of sum and product, or sum and sum of squares.
A1: Obtains \(\frac{5}{8}\) (or 0.625).

(c)
M1: Attempts to find the sum of the new roots, substituting their value from part (a) and \(\alpha\beta\).
M1: Attempts to find the product of the new roots, substituting \(\alpha\beta\).
A1: Formulates the correct equation \(16x^2 - 9x + 4 = 0\) (or any integer multiple thereof, including \(= 0\)).

(a) We first find the determinant of \(\mathbf{M}\):
\[\det(\mathbf{M}) = a(a-2) - (2a-3)(a+1)\]
\[\det(\mathbf{M}) = (a^2 - 2a) - (2a^2 + 2a - 3a - 3)\]
\[\det(\mathbf{M}) = a^2 - 2a - (2a^2 - a - 3)\]
\[\det(\mathbf{M}) = -a^2 - a + 3\]
We are given that \(\det(\mathbf{M}) = -9\):
\[-a^2 - a + 3 = -9 \implies a^2 + a - 12 = 0\]
\[(a+4)(a-3) = 0\]
So, \(a = 3\) or \(a = -4\).

(b) Using the positive value, \(a = 3\), the matrix becomes:
\[\mathbf{M} = \begin{pmatrix} 3 & 3 \\ 4 & 1 \end{pmatrix}\]
Since \(\det(\mathbf{M}) = -9\), the inverse is given by:
\[\mathbf{M}^{-1} = \frac{1}{-9}\begin{pmatrix} 1 & -3 \\ -4 & 3 \end{pmatrix} = \begin{pmatrix} -\frac{1}{9} & \frac{1}{3} \\ \frac{4}{9} & -\frac{1}{3} \end{pmatrix}\]

(c) The area of the transformed triangle is related to the area of the original triangle by:
\[\text{Area}(T') = |\det(\mathbf{M})| \times \text{Area}(T)\]
\[45 = |-9| \times \text{Area}(T) = 9 \times \text{Area}(T)\]
\[\text{Area}(T) = \frac{45}{9} = 5\text{ square units}\]

(a)
M1: Attempts to write an expression for \(\det(\mathbf{M})\) in terms of \(a\).
A1: Simplifies to find \(-a^2 - a + 3\).
M1: Sets their determinant expression equal to \(-9\) to form a quadratic equation and attempts to solve it.
A1: Obtains the correct values \(a = 3\) and \(a = -4\).

(b)
M1: Substitutes \(a=3\) into \(\mathbf{M}\) and attempts to find its inverse using a correct method.
A1: Obtains \(\mathbf{M}^{-1} = -\frac{1}{9}\begin{pmatrix} 1 & -3 \\ -4 & 3 \end{pmatrix}\) or equivalent.

(c)
M1: Recognises and applies the scale factor rule for area: \(\text{Area}(T') = |\det(\mathbf{M})| \times \text{Area}(T)\).
A1: Obtains the correct area of \(5\).

(a) We expand the terms inside the summation:
\[\sum_{r=1}^{n} r(r+3) = \sum_{r=1}^{n} (r^2 + 3r) = \sum_{r=1}^{n} r^2 + 3\sum_{r=1}^{n} r\]
Using the standard summation formulae:
\[\sum_{r=1}^{n} r^2 = \frac{1}{6}n(n+1)(2n+1) \quad \text{and} \quad \sum_{r=1}^{n} r = \frac{1}{2}n(n+1)\]
Substituting these into the equation:
\[\sum_{r=1}^{n} r(r+3) = \frac{1}{6}n(n+1)(2n+1) + 3\left(\frac{1}{2}n(n+1)\right)\]
\[= \frac{1}{6}n(n+1)(2n+1) + \frac{9}{6}n(n+1)\]
Factorising out the common factor \[\frac{1}{6}n(n+1)\]:
\[= \frac{1}{6}n(n+1)\left[(2n+1) + 9\right]\]
\[= \frac{1}{6}n(n+1)(2n+10)\]
\[= \frac{2}{6}n(n+1)(n+5)\]
\[= \frac{1}{3}n(n+1)(n+5)\quad \text{(as required)}\]

(b) Using the result from part (a):
\[\sum_{r=15}^{40} r(r+3) = \sum_{r=1}^{40} r(r+3) - \sum_{r=1}^{14} r(r+3)\]
Let \(f(n) = \frac{1}{3}n(n+1)(n+5)\).
Evaluating for \(n = 40\):
\[f(40) = \frac{1}{3}(40)(41)(45) = 40 \times 41 \times 15 = 24600\]
Evaluating for \(n = 14\):
\[f(14) = \frac{1}{3}(14)(15)(19) = 14 \times 5 \times 19 = 1330\]
Thus:
\[\sum_{r=15}^{40} r(r+3) = 24600 - 1330 = 23270\]

(a)
M1: Expands and splits the summation into \(\sum r^2 + 3\sum r\) and substitutes correct standard formulae.
M1: Factorises out a common factor of at least \(\frac{1}{6}n(n+1)\) or \(\frac{1}{2}n(n+1)\).
A1: Correctly simplifies the remaining factor to \((2n+10)\).
A1*: Reaches the printed answer \(\frac{1}{3}n(n+1)(n+5)\) with no errors in algebraic manipulation.

(b)
M1: Expresses the required sum as the difference between two sums from \(r=1\): \(\sum_{r=1}^{40} - \sum_{r=1}^{14}\).
M1: Substitutes \(n = 40\) and \(n = 14\) correctly into the formula derived in part (a).
A1: Calculates the final answer as \(23270\).

(a) Evaluating the function at the boundaries of the interval:
\[f(0) = 0^3 - 5(0) + 3 = 3\]
\[f(1) = 1^3 - 5(1) + 3 = -1\]
Since \(f(x)\) is a polynomial, it is continuous on \([0, 1]\). Since there is a change of sign between \(f(0) = 3\) and \(f(1) = -1\), there must be at least one root \(\alpha\) in the interval \([0, 1]\).

(b) Using linear interpolation on \([0, 1]\):
\[\frac{x_1 - 0}{1 - x_1} = \frac{f(0)}{-f(1)}\]
\[\frac{x_1}{1 - x_1} = \frac{3}{-(-1)} = 3\]
\[x_1 = 3(1 - x_1) = 3 - 3x_1\]
\[4x_1 = 3 \implies x_1 = 0.75\]
To 3 decimal places, this is \(0.750\).

(c) First, we find the derivative of \(f(x)\):
\[f'(x) = 3x^2 - 5\]
Using \(x_0 = 0.5\):
\[f(0.5) = (0.5)^3 - 5(0.5) + 3 = 0.125 - 2.5 + 3 = 0.625\]
\[f'(0.5) = 3(0.5)^2 - 5 = 0.75 - 5 = -4.25\]
Applying the Newton-Raphson formula:
\[x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}\]
\[x_1 = 0.5 - \frac{0.625}{-4.25} = 0.5 + \frac{5}{34} \approx 0.6470588...\]
To 3 decimal places, the approximation is \(0.647\).

(a)
M1: Evaluates both boundary values \(f(0)\) and \(f(1)\) with at least one correct calculation.
A1: Concludes with a correct statement highlighting both the sign change and continuity of \(f(x)\).

(b)
M1: Attempts to set up a linear interpolation equation (e.g., using ratios or similar triangles).
A1: Solves to find the exact value \(x_1 = 0.75\).
A0.4: Correctly rounds to 3 decimal places as \(0.750\).

(c)
M1: Differentiates \(f(x)\) to get \(f'(x) = 3x^2 - 5\).
M1: Applies the Newton-Raphson formula \(x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}\) using their values for \(f(0.5)\) and \(f'(0.5)\).
A1: Obtains \(0.647\) (rounded correctly to 3 d.p.).

(a) Let \(P(n)\) be the statement that \(\mathbf{M}^n = \begin{pmatrix} 2n + 1 & -4n \\ n & 1 - 2n \end{pmatrix}\).

**Base case:** For \(n = 1\),

\[\mathbf{M}^1 = \begin{pmatrix} 2(1) + 1 & -4(1) \\ 1 & 1 - 2(1) \end{pmatrix} = \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}\]

Since this matches the given matrix \(\mathbf{M}\), \(P(1)\) is true.

**Inductive step:** Assume that \(P(k)\) is true for some positive integer \(k\). That is,

\[\mathbf{M}^k = \begin{pmatrix} 2k + 1 & -4k \\ k & 1 - 2k \end{pmatrix}\]

We must show that \(P(k+1)\) is true, i.e.,

\[\mathbf{M}^{k+1} = \begin{pmatrix} 2(k+1) + 1 & -4(k+1) \\ k+1 & 1 - 2(k+1) \end{pmatrix} = \begin{pmatrix} 2k + 3 & -4k - 4 \\ k + 1 & -2k - 1 \end{pmatrix}\]

Using matrix multiplication:

\[\mathbf{M}^{k+1} = \mathbf{M}^k \mathbf{M} = \begin{pmatrix} 2k + 1 & -4k \\ k & 1 - 2k \end{pmatrix} \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}\]

Multiplying the matrices:

\[\mathbf{M}^{k+1} = \begin{pmatrix} (2k + 1)(3) + (-4k)(1) & (2k + 1)(-4) + (-4k)(-1) \\ k(3) + (1 - 2k)(1) & k(-4) + (1 - 2k)(-1) \end{pmatrix}\]

Simplifying each element:

- Row 1, Column 1: \(3(2k + 1) - 4k = 6k + 3 - 4k = 2k + 3 = 2(k+1) + 1\)
- Row 1, Column 2: \(-4(2k + 1) + 4k = -8k - 4 + 4k = -4k - 4 = -4(k+1)\)
- Row 2, Column 1: \(3k + 1 - 2k = k + 1\)
- Row 2, Column 2: \(-4k - (1 - 2k) = -4k - 1 + 2k = -2k - 1 = 1 - 2(k+1)\)

Thus,

\[\mathbf{M}^{k+1} = \begin{pmatrix} 2(k+1) + 1 & -4(k+1) \\ k+1 & 1 - 2(k+1) \end{pmatrix}\]

Since \(P(1)\) is true, and \(P(k) \implies P(k+1)\), the statement \(P(n)\) is true for all \(n \in \mathbb{Z}^+\) by mathematical induction.

(b) A matrix is non-singular if its determinant is non-zero.

\[\det(\mathbf{M}^n) = (2n + 1)(1 - 2n) - (-4n)(n)\]

\[\det(\mathbf{M}^n) = (1 - 4n^2) + 4n^2 = 1\]

Since \\det(\mathbf{M}^n) = 1
eq 0\) for all \(n \in \mathbb{Z}^+\), the matrix \(\mathbf{M}^n\) is non-singular for all \(n \in \mathbb{Z}^+\).

(a)
B1: Verifies the formula for \(n = 1\).
M1: Assumes formula holds for \(n = k\) and attempts to multiply \(\mathbf{M}^k\) by \(\mathbf{M}\) (or vice versa).
A1: Correct multiplication expression shown.
A1: Correctly simplifies the top-left and bottom-left elements in terms of \(k+1\).
A1: Correctly simplifies the top-right and bottom-right elements in terms of \(k+1\).
A1: Complete and clear conclusion stating that since it is true for \(n=1\), and if true for \(n=k\) then it is true for \(n=k+1\), hence it is true for all \(n \in \mathbb{Z}^+\) by induction.

(b)
M1: Attempts to find the determinant of \(\mathbf{M}^n\) using the formula \(ad - bc\).
A1: Shows that \(\det(\mathbf{M}^n) = 1\) after simplification.
A1: Concludes with a valid reason that since \(\det(\mathbf{M}^n) = 1 \neq 0\), the matrix is non-singular for all \(n \in \mathbb{Z}^+\).

(a) Differentiating the equation of the parabola with respect to \(x\):

\[2y \frac{\mathrm{d}y}{\mathrm{d}x} = 4a \implies \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2a}{y}\]

At the point \(P(ap^2, 2ap)\), the gradient of the tangent is:

\[m_T = \frac{2a}{2ap} = \frac{1}{p}\]

The equation of the tangent at \(P\) is:

\[y - 2ap = \frac{1}{p}(x - ap^2)\]

Multiplying by \(p\):

\[py - 2ap^2 = x - ap^2 \implies py = x + ap^2\]

(b) The equation of the directrix of \(C\) is \(x = -a\).

Substitute \(x = -a\) into the equation of the tangent:

\[py = -a + ap^2 \implies y = \frac{a(p^2 - 1)}{p}\]

Thus, the coordinates of \(Q\) are:

\[Q\left(-a, \frac{a(p^2 - 1)}{p}\right)\]

(c) The focus of the parabola \(C\) is \(S(a, 0)\).

The gradient of \(SP\) is:

\[m_{SP} = \frac{2ap - 0}{ap^2 - a} = \frac{2ap}{a(p^2 - 1)} = \frac{2p}{p^2 - 1}\]

The gradient of \(SQ\) is:

\[m_{SQ} = \frac{\frac{a(p^2 - 1)}{p} - 0}{-a - a} = \frac{a(p^2 - 1)}{p(-2a)} = \frac{p^2 - 1}{-2p} = -\frac{p^2 - 1}{2p}\]

Multiplying the two gradients:

\[m_{SP} \times m_{SQ} = \left(\frac{2p}{p^2 - 1}\right) \times \left(-\frac{p^2 - 1}{2p}\right) = -1\]

Since the product of their gradients is \(-1\), the line \(SQ\) is perpendicular to the line \(SP\).

(a)
M1: Differentiates \(y^2 = 4ax\) implicitly to find an expression for \(\frac{\mathrm{d}y}{\mathrm{d}x}\).
A1: Finds the gradient of the tangent at \(P\) as \(\frac{1}{p}\).
M1: Uses the straight-line equation formula with gradient \(\frac{1}{p}\) and point \((ap^2, 2ap)\).
A1*: Fully simplifies to the given answer \(py = x + ap^2\) with no errors.

(b)
B1: Correctly identifies the equation of the directrix as \(x = -a\).
M1: Substitutes \(x = -a\) into their tangent equation and solves for \(y\).
A1: Correct coordinates of \(Q\left(-a, \frac{a(p^2 - 1)}{p}\right)\).

(c)
B1: Correctly identifies the focus \(S(a, 0)\).
M1: Formulates the gradient of \(SP\) and the gradient of \(SQ\) using \(\frac{y_2 - y_1}{x_2 - x_1}\).
A1: Obtains \(m_{SP} = \frac{2p}{p^2 - 1}\) and \(m_{SQ} = -\frac{p^2 - 1}{2p}\) in simplified forms.
A1: Shows that \(m_{SP} \times m_{SQ} = -1\) and concludes that the lines are perpendicular.

(a) Expanding the term inside the summation:

\[\sum_{r=1}^{n} r(r+3) = \sum_{r=1}^{n} (r^2 + 3r) = \sum_{r=1}^{n} r^2 + 3\sum_{r=1}^{n} r\]

Using the standard summation formulas:

\[\sum_{r=1}^{n} r^2 = \frac{1}{6}n(n+1)(2n+1) \quad \text{and} \quad \sum_{r=1}^{n} r = \frac{1}{2}n(n+1)\]

Substitute these into the equation:

\[\sum_{r=1}^{n} r(r+3) = \frac{1}{6}n(n+1)(2n+1) + \frac{3}{2}n(n+1)\]

Factor out \(\frac{1}{6}n(n+1)\):

\[= \frac{1}{6}n(n+1) [ (2n+1) + 9 ]\]

\[= \frac{1}{6}n(n+1)(2n+10)\]

\[= \frac{2}{6}n(n+1)(n+5) = \frac{1}{3}n(n+1)(n+5)\]

This completes the proof for part (a).

(b) Using the result from part (a):

\[\sum_{r=n}^{2n} r(r+3) = \sum_{r=1}^{2n} r(r+3) - \sum_{r=1}^{n-1} r(r+3)\]

Using the formula with \(2n\) and \(n-1\) respectively:

\[\sum_{r=1}^{2n} r(r+3) = \frac{1}{3}(2n)(2n+1)(2n+5)\]

\[\sum_{r=1}^{n-1} r(r+3) = \frac{1}{3}(n-1)(n)(n+4)\]

Now, subtract the two expressions:

\[\sum_{r=n}^{2n} r(r+3) = \frac{1}{3}(2n)(2n+1)(2n+5) - \frac{1}{3}n(n-1)(n+4)\]

Factor out \(\frac{1}{3}n\):

\[= \frac{1}{3}n [ 2(2n+1)(2n+5) - (n-1)(n+4) ]\]

Expand the terms inside the square brackets:

\[2(2n+1)(2n+5) = 2(4n^2 + 12n + 5) = 8n^2 + 24n + 10\]

\[(n-1)(n+4) = n^2 + 3n - 4\]

So the expression becomes:

\[= \frac{1}{3}n [ (8n^2 + 24n + 10) - (n^2 + 3n - 4) ]\]

\[= \frac{1}{3}n [ 7n^2 + 21n + 14 ]\]

Factor out 7 from the quadratic term:

\[= \frac{7}{3}n(n^2 + 3n + 2)\]

Factorizing \(n^2 + 3n + 2 = (n+1)(n+2)\):

\[= \frac{7}{3}n(n+1)(n+2)\]

This completes the proof for part (b).

(a)
M1: Expands the summand and splits the summation into two parts: \(\sum r^2 + 3\sum r\).
M1: Applies standard formulae for \(\sum r^2\) and \(\sum r\).
M1: Attempts to factorize out common factor \(\frac{1}{6}n(n+1)\) or \(\frac{1}{2}n(n+1)\).
A1*: Fully simplifies to get \(\frac{1}{3}n(n+1)(n+5)\) with no errors shown.

(b)
M1: Writes the sum \(\sum_{r=n}^{2n}\) as \(\sum_{r=1}^{2n} - \sum_{r=1}^{n-1}\).
A1: Writes correct substitution of \(2n\) and \(n-1\) into the formula from (a).
M1: Factors out \(\frac{1}{3}n\) and expands the remaining algebraic terms.
A1: Obtains the correct simplified quadratic expression \(7n^2 + 21n + 14\).
A1*: Correctly factorizes the expression to show \(\frac{7}{3}n(n+1)(n+2)\) with complete intermediate working.

(a) For the quadratic equation \(2x^2 - 3x + 4 = 0\), we have:

\[\alpha + \beta = -\frac{-3}{2} = \frac{3}{2}\]

\[\alpha\beta = \frac{4}{2} = 2\]

We know the identity:

\[\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)\]

Substituting the values of \(\alpha + \beta\) and \(\alpha\beta\):

\[\alpha^3 + \beta^3 = \left(\frac{3}{2}\right)^3 - 3(2)\left(\frac{3}{2}\right)\]

\[\alpha^3 + \beta^3 = \frac{27}{8} - 9\]

\[\alpha^3 + \beta^3 = \frac{27 - 72}{8} = -\frac{45}{8}\]

This completes the proof for part (a).

(b) Let the roots of the new quadratic equation be \(\gamma = \frac{1}{\alpha^3}\) and \(\delta = \frac{1}{\beta^3}\).

The sum of the new roots is:

\[\gamma + \delta = \frac{1}{\alpha^3} + \frac{1}{\beta^3} = \frac{\alpha^3 + \beta^3}{\alpha^3\beta^3} = \frac{\alpha^3 + \beta^3}{(\alpha\beta)^3}\]

Substituting the values from part (a) and \(\alpha\beta = 2\):

\[\gamma + \delta = \frac{-\frac{45}{8}}{2^3} = \frac{-\frac{45}{8}}{8} = -\frac{45}{64}\]

The product of the new roots is:

\[\gamma\delta = \frac{1}{\alpha^3} \times \frac{1}{\beta^3} = \frac{1}{(\alpha\beta)^3}\]

Substituting \(\alpha\beta = 2\):

\[\gamma\delta = \frac{1}{2^3} = \frac{1}{8}\]

The new quadratic equation is given by:

\[x^2 - (\gamma + \delta)x + \gamma\delta = 0\]

\[x^2 - \left(-\frac{45}{64}\right)x + \frac{1}{8} = 0\]

\[x^2 + \frac{45}{64}x + \frac{1}{8} = 0\]

Multiplying the entire equation by 64 to obtain integer coefficients:

\[64x^2 + 45x + 8 = 0\]

This completes the proof for part (b).

(a)
B1: States correct values of \(\alpha + \beta = \frac{3}{2}\) and \(\alpha\beta = 2\).
M1: Recalls or derives the correct identity \(\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)\).
M1: Substitutes their values of sum and product into the identity.
A1*: Obtains the exact value \(-\frac{45}{8}\) with no errors.

(b)
M1: Expresses the sum of the new roots \(\frac{1}{\alpha^3} + \frac{1}{\beta^3}\) with a common denominator of \(\alpha^3\beta^3\).
A1: Calculates the correct sum of the new roots as \(-\frac{45}{64}\).
M1: Calculates the correct product of the new roots as \(\frac{1}{8}\).
M1: Forms a quadratic equation of the form \(x^2 - S x + P = 0\) with their sum and product.
A1*: Multiplies by 64 to obtain the correct final equation \(64x^2 + 45x + 8 = 0\).