题目

Problem

(a) Find the first 4 terms of the binomial expansion, in ascending powers of $x$ , of

(8-3x)^{-\frac{1}{3}}\qquad |x|<\frac{8}{3}

giving each coefficient as a simplified fraction.

(4)

(b) Use the answer from part (a) with $x=\dfrac{2}{3}$ to find a rational approximation to $\sqrt[3]{6}$ .

(2)

题目中文翻译

(a) 求

(8-3x)^{-\frac{1}{3}}\qquad |x|<\frac{8}{3}

按 $x$ 的升幂展开时的前 4 项，并将每个系数化为最简分数。

(b) 利用 (a) 的结果并取 $x=\dfrac{2}{3}$ ，求 $\sqrt[3]{6}$ 的一个有理近似值。

解答

Rewrite the expression:

(8-3x)^{-\frac13} =8^{-\frac13}\left(1-\frac{3x}{8}\right)^{-\frac13}

Since

8^{-\frac13}=\frac12

we have

(8-3x)^{-\frac13} =\frac12\left(1-\frac{3x}{8}\right)^{-\frac13}

Use the binomial expansion

(1+u)^n =1+nu+\frac{n(n-1)}{2}u^2 +\frac{n(n-1)(n-2)}{6}u^3+\cdots

Here

n=-\frac13,\qquad u=-\frac{3x}{8}

\begin{aligned} \left(1-\frac{3x}{8}\right)^{-\frac13} &=1+\left(-\frac13\right)\left(-\frac{3x}{8}\right) \\ &\quad+\frac{\left(-\frac13\right)\left(-\frac43\right)}{2} \left(-\frac{3x}{8}\right)^2 \\ &\quad+\frac{\left(-\frac13\right)\left(-\frac43\right)\left(-\frac73\right)}{6} \left(-\frac{3x}{8}\right)^3+\cdots \end{aligned}

This simplifies to

\left(1-\frac{3x}{8}\right)^{-\frac13} =1+\frac{x}{8}+\frac{x^2}{32}+\frac{7x^3}{768}+\cdots

Therefore

(8-3x)^{-\frac13} =\frac12+\frac{x}{16}+\frac{x^2}{64}+\frac{7x^3}{1536}+\cdots

Hence the first four terms are

\boxed{ \frac12+\frac{x}{16}+\frac{x^2}{64}+\frac{7x^3}{1536} }

When

x=\frac23

we have

8-3x=8-3\left(\frac23\right)=6

So the expansion gives an approximation to

6^{-\frac13}

Substitute $x=\dfrac23$ into the expansion:

\begin{aligned} 6^{-\frac13} &\approx \frac12+\frac{1}{16}\left(\frac23\right) +\frac{1}{64}\left(\frac23\right)^2 +\frac{7}{1536}\left(\frac23\right)^3 \\ &=\frac12+\frac1{24}+\frac1{144}+\frac7{5184} \\ &=\frac{2851}{5184} \end{aligned}

Thus

\frac{1}{\sqrt[3]{6}}\approx\frac{2851}{5184}

Taking the reciprocal,

\sqrt[3]{6}\approx\frac{5184}{2851}

Therefore a rational approximation is

\boxed{\frac{5184}{2851}}