Find the first three terms, in ascending powers of $x$, in the binomial expansion of $(1 + 6x)^{rac{1}{3}}$ - AQA - A-Level Maths Pure - Question 5 - 2019 - Paper 3

From part (a), we found the expansion of (1 + 6x)^{rac{1}{3}} as:

$1 + 2x - 8x^2$

To approximate rac{1}{ oot{18}}, we can set $6x$ equal to $-1$ (since $(1 + 6x)$ should be close to $1$ for small $x$). Thus, we have:

$6x = -1 \implies x = -\frac{1}{6}$

Substituting into the expansion gives:

$1 + 2(-\frac{1}{6}) - 8(-\frac{1}{6})^2 = 1 - \frac{1}{3} - \frac{8}{36}$

This simplifies to:

$1 - \frac{1}{3} - \frac{2}{9} = \frac{27 - 9 - 6}{27} = \frac{12}{27} \approx 0.4444$

Thus, the approximation to rac{ oot{1}{18}} is approximately $0.4444$.

Substituting $x = \frac{1}{2}$ into the expansion from part (a):

$1 + 2(\frac{1}{2}) - 8(\frac{1}{2})^2 = 1 + 1 - 2 = 0$

This results in zero, and since we are looking for an approximation to rac{ oot{4}{3}}, which is greater than zero, this substitution is inappropriate. Moreover, the expansion is only valid for $|6x| < 1$; substituting $x = \frac{1}{2}$ gives:

$6(\frac{1}{2}) = 3$

which violates this condition.