Compute $\sqrt[7]{0.999}$ to three decimal places.(From Gelfand's Algebra text.) After a brief introduction to roots(imaginary numbers were not introduced yet), this question is asked. I am apparently expected to find the answer using elementary algebraic manipulation. I have tried playing around with the problem and observed 
$\sqrt[7]{0.999} = {(1-\frac{1}{1000}})^\frac{1}{7} = (\frac{(3)^3(37)}{10^3})^\frac{1}{7}$
But do not get how to solve the problem.Any helpful ideas will be appreciated.
PS: Inequalities and binomial series were not yet introduced in the text.I have only a rudimentary knowledge of maths, so keep the answers simple and within the scope of elementary algebra.
I have edited this problem,adding context. Do not penalize previous answers which used calculus. :)
 A: Rob Arthan wrote in a comment: "What is clear is that Gelfand expected his readers to be prepared to do the long multiplications that show that $0.9995^3<0.999$ (no calculus required!)".
To fill in this gap, I'd like to note that we don't actually have to do the multiplication. Note that $0.9995 = 1 - 0.005$.  Therefore we have
$0.9995^3 = (1-0.005)^3 = 1 - 3 \times 0.005 + 3 \times 0.005^2 - 0.005^3$
by the binomial theorem.  (If one has not seen the binomial theorem, it's not hard to just expand $(1-x)^3$ "by hand" and plug in $x = 0.005$.)  But $3 \times 0.005^2 < 0.005$.  Thus
$1 - 3 \times 0.005 + 3 \times 0.005^2 - 0.005^3 < 1 - 3 \times 0.005 + 0.005 - 0 = 0.999$
which proves the desired result.
A: For a general answer, I would use the Taylor expansion of $(1+x)^a$:
$$(1+x)^a = 1+ax+\frac{a(a-1)}{2}x^2+ \cdots + \frac{a(a-1)\cdots(a+1-n)}{n!}x^n+\cdots$$
Here with $a=1/7$ and $x=-0.001$
A: The taylor series of $\sqrt[a]{1-x}$
$$ y=1-\frac{x}{a}-.-.-.$$ 
at $x=0.001$ and a=7
$$y=1-0.001/7$$ 
$$\sqrt[7]{0.999}=1-\frac{1}{7000}=0.9998571$$
A: For a precalculus solution, you presumably can't use the Taylor series.  You can note that $0.9995^7 \lt 0.997$ (shown here by Alpha), so $\sqrt[7]{0.999} \gt 0.9995$ and to three decimal places it will round to $1.000$
