(a) Use the binomial theorem to expand (8-3x)^{3/2}, |x| < rac{8}{3}, up to and including the term in x^{3}, giving each term as a simplified fraction - Edexcel - A-Level Maths Pure - Question 4 - 2008 - Paper 8

To expand ( (8-3x)^{3/2} ) using the binomial theorem:

1. Identify Constants:

   • Let ( a = 8 ) and ( b = -3x ), then we have:
   $$(a + b)^{n} = \sum_{k=0}^{n} {n \choose k} a^{n-k} b^{k}$$

2. Apply Values:

   • In our case, ( n = \frac{3}{2} ):
   $$(8 + (-3x))^{3/2}$$
   • Therefore, the expansion will be:
   $$\sum_{k=0}^{3} {\frac{3}{2} \choose k} (8)^{\frac{3}{2}-k} (-3x)^{k}$$

3. Calculate Terms:

   • Now we will compute each term up to ( k=3 ):

   • When k = 0:

   $${\frac{3}{2} \choose 0} (8)^{\frac{3}{2}} (-3x)^{0} = 8^{\frac{3}{2}} = 8 \cdot \sqrt{8} = 16\sqrt{2}$$
   • When k = 1:
   $${\frac{3}{2} \choose 1} (8)^{\frac{3}{2}-1} (-3x) = \frac{3}{2} \cdot (8)^{\frac{1}{2}} (-3x) = -12 \cdot 2\sqrt{2} x = -24\sqrt{2} x$$
   • When k = 2:
   $${\frac{3}{2} \choose 2} (8)^{\frac{3}{2}-2} (-3x)^{2} = \frac{3/2 \cdot 1/2}{2} (4)(9x^2) = \frac{27}{8} \cdot 4 = \frac{27}{2} x^2$$
   • When k = 3:
   $${\frac{3}{2} \choose 3} (8)^{\frac{3}{2}-3} (-3x)^{3} = 0 ext{ (as it exceeds the power)}$$

4. Final Expansion:

   • Putting all terms together, the expansion up to ( x^{3} ) is:
   $$16\sqrt{2} - 24\sqrt{2} x + \frac{27}{2} x^2$$

   Therefore, simplified, the final answer is:

   $$16\sqrt{2} - 24\sqrt{2} x + \frac{27}{2} x^2$$