Quickly simplify $(\sqrt{3}+1)^6+(\sqrt{3}-1)^6$

Question

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.

Answer 1

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 $$(a+b)(a^2-ab+b^2)=8(2\times28-4)=416$$

Answer 2

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

$$m^6+n^6=8^3-12\times 8=416.$$

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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:

$$\begin{align}m^6+n^6=\left(m^2\right)^3+\left(n^2\right)^3=\left(m^2+n^2\right)^3-3\left(mn\right)^2\left(m^2+n^2\right).\end{align}$$

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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

$$\begin{align}\left(\sqrt 3+1\right)^6+\left(\sqrt 3-1\right)^6&=8(64-12)\\ &=416.\end{align}$$

Answer 3

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}