If $$a-\frac{1}{a}=1$$, then $$a^2+\frac{1}{a^2}=?$$
GivenÂ
$$a-\frac{1}{a}=1$$
the
$$^{a^2+\frac{1}{a^2}=?}$$
$$a-\frac{1}{a}=1$$ squaring both side we getÂ
$$a^2+\frac{1}{a^2}-2\cdot a\cdot\frac{1}{a}=1^2$$
so we get $$a^2+\frac{1}{a^2}=3$$
3 is the answer
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