Use Cauchy product to find a power series represenitation of $1 \over {(1-x)^3}$ 
Use Cauchy product to find a power series represenitation of
  $$1 \over {(1-x)^3}$$ which is valid in the interval $(-1,1)$.

Is it right to use the product of $1 \over {1-x}$ and $1 \over {(1-x)^2}$ if I know thier expantion (HOW).. Is there another way?

This is most easily done with the binomial theorem and the identity
$$
\binom{-n}{k}=(-1)^k\binom{n+k-1}{k}
$$
which gives
$$
\begin{align}
\frac1{(1-x)^3}
&=\sum_{k=0}^\infty\binom{-3}{k}(-x)^k\\
&=\sum_{k=0}^\infty\binom{k+2}{k}x^k\\
&=\sum_{k=0}^\infty\frac{(k+2)(k+1)}{2}x^k
\end{align}
$$

To use the Cauchy products, we would use the $n=1$ version of the series above
$$
\frac1{1-x}=\sum_{k=0}^\infty x^k\tag{1}
$$
The Cauchy product method says that
$$
\left(\sum_{k=0}^\infty a_k x^k\right)\left(\sum_{k=0}^\infty b_k x^k\right)
=\sum_{k=0}^\infty\left(\sum_{j=0}^k a_jb_{k-j}\right)x^k\tag{2}
$$
$(2)$ applied to $(1)$ twice gives
$$
\begin{align}
\frac1{(1-x)^2}
&=\sum_{k=0}^\infty\left(\sum_{j=0}^k1\cdot1\right)x^k\\
&=\sum_{k=0}^\infty(k+1)x^k\tag{3}
\end{align}
$$
$(2)$ appplied to $(1)$ and $(3)$ yields
$$
\begin{align}
\frac1{(1-x)^3}
&=\sum_{k=0}^\infty\left(\sum_{j=0}^k(j+1)\cdot1\right)x^k\\
&=\sum_{k=0}^\infty\frac{(k+2)(k+1)}{2}x^k\tag{4}
\end{align}
$$
