<s>[INST] If $a^2-6a-1=0$, find the value of $a^2 +\dfrac {1}{a^2}$ If $a^2-6a-1=0$, find the value of $a^2 +\dfrac {1}{a^2}$
My Attempt:
$$a^2+\dfrac {1}{a^2}=(a+\dfrac {1}{a})^2 - 2a.\dfrac {1}{a}$$
$$a^2+\dfrac {1}{a^2}=(a+\dfrac {1}{a})^2 - 2$$
How do I proceed further?
 [/INST] You have $$a-\frac 1a=6$$ so squaring gives $$a^2-2+\frac{1}{a^2}=36$$ so...?
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