If a is small, then show that $[1/(1+a)]^3$ is nearly equal to $1-3a$.

If a is small, then show that $$\dfrac{1}{(1+a)^3}$$ is nearly equal to $$(1-3a)$$. Also show that if $$0, then the error$$<0.0007$$.

Can we solve this without using calculus?

If so can anyone help me with the solution?

We know that $$(1+a)^3(1-a)^3=(1-a^2)^3$$ Therefore $$\frac{1}{(1+a)^3}=(1-a)^3/(1-a^2)^3\approx (1-3a+3a^2-a^3)\approx1-3a$$ Now for small a we can neglect $$a^2$$ and$$a^3$$ that's because higher powers of a get very small when compared to 1 or a when a is small Example $$a=0.001$$ then$$a^2=0.00001$$ and$$a^3=0.0000001$$ which are very small . $$Q.E.D.$$
Suppose $$a$$ is small. Then $$(1+a)^x \approx 1+xa$$ holds and is known as the Binomial approximation. Thus $$\frac{1}{(1+a)^3} = (1+a)^{-3} \approx 1+(-3)a = (1-3a)$$.
To find the error, note that as $$a \rightarrow 0$$, $$\frac{1}{(1+a)^3} \rightarrow 1$$ and $$(1-3a) \rightarrow 1$$. When $$a=0.01$$, the error between the two expressions is $$|\frac{1}{(1+a)^3} - (1-3a)| = |\frac{1}{[1+(0.01)]^3} - [1-3(0.01)]| \approx (0.9706) - (0.9700) = 0.0006$$, which is less than $$0.0007$$. Noting that the difference in the expressions is monotonically increasing in the interval $$0, we can say that the error is thus less than $$0.0007$$.
\begin{align} \frac1{(1+a)^3}-(1-3a) &=\frac{6+8a+3a^2}{(1+a)^3}\cdot a^2\\[6pt] &\le\left\{\begin{array}{} 22a^2&\text{if }a\ge-\frac12\\ 6a^2&\text{if }a\ge0 \end{array}\right. \end{align} since $$\frac{6+8a+3a^2}{(1+a)^3}=\frac3{1+a}+\frac2{(1+a)^2}+\frac1{(1+a)^3}$$ is decreasing for $$a\gt-1$$.
If $$0\le a\le0.01$$, then the error is at most $$0.0006$$.