Quickly simplify $(\sqrt{3}+1)^6+(\sqrt{3}-1)^6$ Here's my question:
Simplify $(\sqrt{3}+1)^6+(\sqrt{3}-1)^6$.
I'm aware that I can just use binomial theorem to expand each of the terms individually and then just cancel/add/subtract the like terms however I'm wondering whether there is a quicker way to solve this question.
 A: I don't know if this is any quicker, but you could try:
Let $$a=(\sqrt{3}+1)^2=4+2\sqrt{3}$$
And let $$b=(\sqrt{3}-1)^2=4-2\sqrt{3}$$
Then $$a^2=28+16\sqrt{3}\implies b^2=28-16\sqrt{3}$$
And $$ab=4$$
Then the required expression is $$(a+b)(a^2-ab+b^2)=8(2\times28-4)=416$$
A: The fastest ( and the least error-prone), in my opinion, would be to calculate literally: set $a=\sqrt 3$; we have:
\begin{align}
(a+1)^6+(a-1)^6&=\phantom{+}a^6+6a^5+15a^4+20a^3+15a^2+6a+1 \\
&\phantom{=}+a^6-6a^5+15a^4-20a^3+15a^2-6a+1 \\
&=\color{red}{2(a^6+15a^4+15a^2+1)}\\
&=\color{red}{2(27+15\cdot9+15\cdot3+1)=416}
\end{align}
A: One of the quick methods without using the binomial theorem can be constructed as follows:
Let $\sqrt 3
+1=m,\thinspace \sqrt 3-1=n$, then we have
$$\begin{cases}m^2+n^2=8\\mn=2\end{cases}$$
Then using the formula,
$$\begin{align}m^6+n^6=\left(m^2+n^2\right)^3-3\left(mn\right)^2\left(m^2+n^2\right)\tag 1\end{align}$$
we get
$$m^6+n^6=8^3-12\times 8=416.$$

Explanation: $(1)$
I used the following well-known formula:
$$\begin{align}m^3+n^3=(m+n)^3-3mn(m+n)\end{align}$$
Then, we can derive the required equality:
$$\begin{align}m^6+n^6=\left(m^2\right)^3+\left(n^2\right)^3=\left(m^2+n^2\right)^3-3\left(mn\right)^2\left(m^2+n^2\right).\end{align}$$

Small Supplement:
Based on the formula $(1)$, we can also use the following identity:
$$m^6+n^6=\left(m^2+n^2\right)\left(\left(m^2+n^2\right)^2-3(mn)^2\right)$$
where, $m^2+n^2=8$ and $mn=2$.
Thus, we have
$$\begin{align}\left(\sqrt 3+1\right)^6+\left(\sqrt 3-1\right)^6&=8(64-12)\\
&=416.\end{align}$$
