<s>[INST] If $a+\frac1a=\sqrt3$ then $a^4+\frac1{a^4}=\ ?$ 
If $a+\frac1a=\sqrt3$ then $a^4+\frac1{a^4}=\ ?$

Can someone please explain to me how to solve this?
because I tried everything I know and it didn't work.
P.S: I'm in 8th grade so no quadratic formula.
 [/INST] Expand $\left(a + \frac{1}{a}\right)^4$ by using foil, repeatedly using the famous result $(a+b)^2 = a^2 + 2ab + b^2$ (which is best in this case) or as I've done, you can use the binomial theorem assisted by pascal's triangle:
$$\left(a + \frac1{a}\right)^4 = \left(\sqrt{3}\right)^4\\
\\ a^4 + 4\cdot a^3\frac{1}{a} + 6\cdot a^2\left(\frac{1}{a}\right)^2 + 4\cdot a\left(\frac{1}{a}\right)^3 + \left(\frac{1}{a}\right)^4 = 3^{4/2}\\
a^4 + 4a^2 + 6 + \frac{4}{a^2} + \frac{1}{a^4} = 3^2\\
a^4 + \frac{1}{a^4} + 4\left(a^2 + \frac{1}{a^2}\right) = 9 - 6\\
a^4 + \frac{1}{a^4} + 4\left(\Big(a + \frac{1}{a}\Big)^2 - 2\, a\frac{1}{a}\right) = 3\\
a^4 + \frac{1}{a^4} + 4\left((\sqrt{3})^2 - 2\right) = 3\\
\therefore \quad a^4 + \frac{1}{a^4} = 3 - 4\cdot(3-2) = -1
$$
What you see above is a gut response to this question, if you've paid attention, you will notice my usage of $a^2 + b^2 = (a+b)^2 - 2ab$ and the important substitution of $a + \frac{1}{a} = \sqrt{3}$
There are loads of ways to do algebra. The results matter more than the method. Intuition matters more than anything else.
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