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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.

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    $\begingroup$ You could determine the minial polynomials of the bases and try to find idendities between them, but this is barely easier than just calculating the expression. This could however be useful , if the exponents were significantly higher. $\endgroup$
    – Peter
    Oct 26 at 9:42
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    $\begingroup$ Hint: Let $x_n=(\sqrt{3}+1)^n+(-\sqrt{3}+1)^n$. Then $x_{n+2}=2x_{n+1}+2x_{n}$. $\endgroup$
    – lhf
    Oct 26 at 9:58
  • $\begingroup$ See also oeis.org/A080040 $\endgroup$
    – lhf
    Oct 26 at 10:02
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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 $$(a+b)(a^2-ab+b^2)=8(2\times28-4)=416$$

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  • $\begingroup$ Damn yeah that makes so much more sense considering a^3-b^3. Thanks mate $\endgroup$ Oct 26 at 10:21
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    $\begingroup$ You're welcome. Nevertheless it's worth pointing out that you only need to do the first bracket expansions using Pascal's Triangle/Binomial Theorem, because then you can automatically write down the second one which will be the same but with a sign change before the surd term. All the surds then cancel. $\endgroup$ Oct 26 at 11:23
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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}

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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

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

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