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2004 Paper 1 Q1
D: 1500.0 B: 1516.0
surds binomial expansion cube root algebraic manipulation quadratic in disguise integer identification substitution

  1. Express \(\left(3+2\sqrt{5} \, \right)^3\) in the form \(a+b\sqrt{5}\) where \(a\) and \(b\) are integers.
  2. Find the positive integers \(c\) and \(d\) such that \(\sqrt[3]{99-70\sqrt{2}\;}\) = \(c - d\sqrt{2} \,\).
  3. Find the two real solutions of \(x^6 - 198 x^3 + 1 = 0 \,\).

Solution:

  1. \begin{align*} (3+2\sqrt{5})^3 &= 3^3 + 3 \cdot 3^2 \cdot 2\sqrt{5} + 3 \cdot 3 \cdot (2 \sqrt{5})^2 + (2\sqrt{5})^3 \\ &= 27 + 180 + (54+40)\sqrt{5} \\ &= 207 + 94\sqrt{5} \end{align*}
  2. \begin{align*} && (c-d\sqrt{2})^3 &= c^3+6cd -(3c^2d+2d^3)\sqrt{2} \\ \Rightarrow && 99 &= c(c^2+6d^2) \\ && 70 &= d(3c^2+2d^2) \\ \Rightarrow && c & \mid 99, d \mid 70 \\ && c &= 3, d = 2 \end{align*}
  3. \begin{align*} && 0 &= x^6 - 198x^3 + 1 \\ \Rightarrow && 0 &= (x^3-99)^2+1-99^2 \\ \Rightarrow && x^3 &= 99 \pm \sqrt{99^2-1} \\ &&&= 99 \pm 10 \sqrt{98} \\ &&&= 99 \pm 70 \sqrt{2} \\ \Rightarrow && x &= 3 \pm 2 \sqrt{2} \end{align*}

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