Question
If $a \neq 0$ and $a-\frac{1}{a}=4 ;$ find $:\left(a^4+\frac{1}{a^4}\right)$
✓
Answer
$\left(a-\frac{1}{a}\right)^2=a^2+\frac{1}{a^2}-2$
$ \Rightarrow a^2+\frac{1}{a^2}=\left(a-\frac{1}{a}\right)^2+2$
$ \Rightarrow a^2+\frac{1}{a^2}=(4)^2+2 \quad\left[\because a-\frac{1}{a}=4\right]$
$ \Rightarrow a^2+\frac{1}{a^2}=18 \ldots(1)$
We know that,
$a^4+\frac{1}{a^4}= \left(a^2+\frac{1}{a^2}\right)^2-2$
$ ={ }^{\prime}(18)^{\wedge} 2-2 \quad[$ From$(1)]$
$ =324-2$
$\Rightarrow a^4+\frac{1}{a^4}$
$=322$
$ \Rightarrow a^2+\frac{1}{a^2}=\left(a-\frac{1}{a}\right)^2+2$
$ \Rightarrow a^2+\frac{1}{a^2}=(4)^2+2 \quad\left[\because a-\frac{1}{a}=4\right]$
$ \Rightarrow a^2+\frac{1}{a^2}=18 \ldots(1)$
We know that,
$a^4+\frac{1}{a^4}= \left(a^2+\frac{1}{a^2}\right)^2-2$
$ ={ }^{\prime}(18)^{\wedge} 2-2 \quad[$ From$(1)]$
$ =324-2$
$\Rightarrow a^4+\frac{1}{a^4}$
$=322$
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