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.
 A: We have these identities. (Do you see why these are true? Hint: FOIL. Or, if you learned $``{(a+b)^2=a^2+2ab+b^2}
``$, use that.)
$$\left(a+\frac1a\right)^2=\left(a^2+\frac1{a^2}\right)+2$$
$$\left(a^2+\frac1{a^2}\right)^2=\left(a^4+\frac1{a^4}\right)+2$$
Now, plug in $(a+1/a)=\sqrt3$.
$$\left(\sqrt3\right)^2=\left(a^2+\frac1{a^2}\right)+2$$
So $3=(a^2+1/a^2)+2$, and $a^2+1/a^2=1$.
$$(1)^2=\left(a^4+\frac1{a^4}\right)+2$$
So $1=(a^4+1/a^4)+2$, or $a^4+1/a^4=-1$ and we have solved the problem.
A: Apply $p^2+q^2=(p+q)^2-2pq$ 
$$a^4+\frac1{a^4}=\left(a^2\right)^2+\left(\frac1{a^2}\right)^2$$
and on $$a^2+\frac1{a^2}=a^2+\left(\dfrac1a\right)^2$$
A: $$3=\left(a+\frac 1a\right)^2=a^2+\frac1{a^2}+2$$
so
$$1=a^2+\frac1{a^2}$$
Make the same manipulation again and you get it.
But it seems that all that is impossible for a real number $a$...
A: $$\left(a^n+\frac1{a^n}\right)\left(a+\frac1a\right)=\left(a^{n+1}+\frac1{a^{n+1}}\right)+\left(a^{n-1}+\frac1{a^{n-1}}\right)$$
$$\implies a^{n+1}+\frac1{a^{n+1}}=\left(a^n+\frac1{a^n}\right)\left(a+\frac1a\right)-\left(a^{n-1}+\frac1{a^{n-1}}\right)$$
If $T_m=a^m+\dfrac1{a^m},$
$$T_{n+1}=T_n\cdot T_1-T_{n-1}$$
We have $T_1=\sqrt3,T_0=2$ as $a\ne0$
A: 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.
