Process for $(k+1)^3$? I've mentioned in previous questions that I have a hard time with simplifying algebraic equations. 
For this equation, I assumed we would first put the equation into easier to read form:
$(k+1)(k+1)(k+1)$
Then apply FOIL to the first two:
EDIT: $(k^2+2k+1)(k+1)$
Although this is where I'm stuck. Any help is appreciated. 
With all your help my answer is:
$k^3+3k^2+3k+1$
Thanks. 
 A: You just multiply each pair of terms:
$$(k^2+2k+2)(k+1)=k^2\cdot k + k^2\cdot 1 + 2k\cdot k + 2k\cdot 1 + 2 \cdot k + 2 \cdot 1 $$
Of course you can also use the binomial theorem, but I'd try to master the basics first.
A: Use the distributive law (after correcting your first multiplication):
$$(k^2+2k+1)(\color{red}{k}+\color{blue}{1})=(k^2+2k+1)\cdot\color{red}{k}+(k^2+2k+1)\cdot\color{blue}{1}\;.$$
Now $(k^2+2k+1)\cdot k=k^3+2k^2+k$, and $(k^2+2k+1)\cdot 1=k^2+2k+1$, so all that’s left is to add the two polynomials:
$$\begin{align*}
(k^3+2k^2+k)+(k^2+2k+1)&=k^3+(2k^2+k^2)+(k+2k)+1\\
&=k^3+3k^2+3k+1\;.
\end{align*}$$
You can also organize this calculation like a pencil-and-paper multiplication:
$$\begin{array}{r}
&&k^2&+&2k&+&1\\
&&&&k&+&1\\ \hline
&&k^2&+&2k&+&1\\
k^3&+&2k^2&+&k\\ \hline
k^2&+&3k^2&+&3k&+&1
\end{array}$$
A: You can write $(k^2+2k+1)(k+1)=k(k^2+2k+1)+1(k^2+2k+1)$ then distribute to get $k(k^2+2k+1)+1(k^2+2k+1)=k^3+2k^2+k+k^2+2k+1$ then collect like terms to get $k^3+3k^2+3k+1$.  This is more general than FOIL, which only handles the product of two items of two terms each.
A: You can apply a similar procedure and distribute the two binomial terms over the tri-nomial:
$$ \begin{align*}(k^2 + 2k + 1) (k+1) 
 &=  (k^2 + 2k^\hphantom{2} + 1) k +  (k^2 + 2k + 1) 1\\
 &=  \hphantom{(}k^3 + 2k^2 + k\hphantom{1)} +  \hphantom{(}k^2 + 2k + 1 \\
 &=  k^3 + 3k^2 + 3k + 1 \end{align*}$$
Note:


*

*You have a mistake in the question: $(k+1)^2 = (k^2 + 2k + 1)$

A: Multiplication is distributive over addition, so:
$$(k^2+2k+2)(k+1)=(k^2+2k+2)(k)+ (k^2+2k+2)(1) $$
$$=(k^3+2k^2+2k)+(k^2+2k+2)$$
Now, sum up like terms:
$$=k^3+3k^2+4k+2$$
Hope that helps.
A: You might want to learn how this triangle works:
$$
1\\
1\,1\\
1\,2\,1\\
1\,3\,3\,1\\
1\,4\,6\,4\,1\\
1\,5\,10\,10\,5\,1
$$
The $n^{th}$ line (first one is zero) gives the coefficient of $x^{k}y^{n-k}$ in $(x+y)^n$
Since you had $(k+1)^3$, the coefficient are gonna be $1,3,3,1$ and now we have to figure out how to work the $k$. 
