$\left(\sqrt{8}+\sqrt{2}\right)^2$ = 18 why?? I would like to now why 
$$\left(\sqrt{8}+\sqrt{2}\right)^2$$
is equal to $18$? Please provide me
the proccess. Thank you.
 A: Hint:
$$\begin{align}
\sqrt{8}+\sqrt{2}&=2\sqrt{2}+\sqrt{2}\\
&=(2+1)\sqrt{2}\\
&=3\sqrt{2}
\end{align}$$
Thus,
$$\left(\sqrt{8}+\sqrt{2}\right)^2=???$$
A: Take out the common factor of $\sqrt2$.
$$
\begin{align*}
(\sqrt8+\sqrt2)^2
&=(2\sqrt2+\sqrt2)^2\\
&=\sqrt2^2(2+1)^2\\
&=2\times3^2\\
&=\cdots\\
\end{align*}$$
A: Hint: $\sqrt{8 \cdot 2} = \sqrt{16} = 4$.
A: $$ (\sqrt{8}+\sqrt{2})^2 = (\sqrt{8})^2+2\sqrt{2}\sqrt{8}+(\sqrt{2})^2=8+2\sqrt{2\cdot 8}+2 = 10 + 2 \cdot 4 = 18$$
A: Hint: $(a+b)^2 = a^2 + b^2 + 2ab$
A: It does look counter-intuitive, doesn't it? But if you step through the multiplication, I think it will become crystal clear to you. First of all, $$(\sqrt{8} + \sqrt{2})^2 = (\sqrt{8} + \sqrt{2})(\sqrt{8} + \sqrt{2})$$, right? Applying FOIL (First, Outer, Inner, Last), we get $$\sqrt{8} \sqrt{8} = 8$$ $$\sqrt{8} \sqrt{2} = 4$$ $$\sqrt{2} \sqrt{8} = 4$$ $$\sqrt{2} \sqrt{2} = 2$$ And then $8 + 4 + 4 + 2 = 18$.
A: $$\left(\sqrt{8}+\sqrt{2}\right)^2=\left(2\sqrt{2}+\sqrt{2}\right)^2=\left(3\sqrt{2}\right)^2=9\times2=18$$
A: Hint :
$$
(a+b)^2=(\color{red}a+\color{blue}b)(a+b)=\color{red}a(a+b)+\color{blue}b(a+b)=a^2+2ab+b^2.
$$
