Expanding $(x-2)^3$ I was trying to expand $(x-2)^3$. This is what I did


*

*Expanded the term so $(x-2)(x-2)(x-2)$

*Multiplied the first term and second term systematically through each case


The answer I got did not match the one at the back of the book, can someone show me how to do this please?
 A: If you are not familiar with the Binomial Theorem, you can expand $(x - 2)^3$ as follows:
\begin{align*}
(x - 2)^3 & = (x - 2)[(x - 2)(x - 2)]\\
          & = (x - 2)[x(x - 2) - 2(x - 2)]\\
          & = (x - 2)(x^2 - 2x - 2x + 4)\\
          & = (x - 2)(x^2 - 4x + 4)\\
          & = x(x^2 - 4x + 4) - 2(x^2 - 4x + 4)\\
          & = x^3 - 4x^2 + 4x - 2x^2 + 8x - 8\\
          & = x^3 - 6x^2 + 12x - 8
\end{align*}
The key idea here is that we can only multiply two factors at once, so we first find the product of two of the factors, then multiply the result by the third factor.
A: Pascal's Triangle (below) and the Binomial Theorem are key elements in solving expansions such as you are working with. From the linked page, a generalised formula is:
$$(x+y)^n = \underset{k=0}{\overset{n}\sum}\binom{n}{k}x^{n-k}y^k$$
So, for a cubic;
$$(x+y)^3=x^3 + 3x^2y + 3xy^2 + y^3$$
In your case, the $y$ term can be replaced with $-2$
You will note that the coefficients on the cubic match the 4th row of Pascal's Triangle below.
Pascal's Triangle

A: Make use of the identity: $$(x-y)^3=x^3-3x^2y+3xy^2-y^3.$$
Letting $y=2$, we obtain: $$(x-2)^3=x^3-6x^2+12x-8.$$
