Please explain the formula for the sum of the cubes and the difference: $a^3 - b^3$ and $a^3 + b^3$? I have not yet fully mastered the formula for accelerated multiplication. I was able to understand everything except the last two. I would like to deal with them. Why have : $$ a^3 + b^3 = (a+b)(a^2 -ab + b^2)^* $$ and $$ a^3 - b^3 = (a-b)(a^2 + ab + b^2) $$ Can someone paint me a formula in detail. Why is this the way it works.

There is a very general factorisation formula, once taught in high school:
$$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\dots+a^{n-k}b^k+\dots+ab^{n-2}+b^{n-1}).$$
(To memorise it, the second factor is the sum of all monomials in $a$ and $b$ of total degree $n-1$.)
The simplest proof consists in showing first by induction the particular case
$$1-x^n=(1-x)\cdot(1+x+x^2+\dots+x^n), $$
then, for the general case, we may suppose that $a\ne 0$ and we set $x=\frac ba$, so that
\begin{align}
a^n-b^n&=a^n(1-x^n)=a^n(1-x)(1+x+x^2+\dots+x^{n-1})\\
&=(a-ax)(a^{n-1}+a^{n-1}x+a^{n-1}x^2+\dots+a^{n-1}x^{n-1})
\end{align}
and observing that, for each $k$, $\;a^{n-1}x^k=a^{n-1-k}b^k$.
For the sum of powers, there is an analogous formula for $n$ odd, with an alternating sum:
$$a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b+\dots+(-1)^{k-1}a^{n-k}b^k+\dots-ab^{n-2}+b^{n-1}).$$
