Question : If $x=\sqrt2+1$, then the value of $x^{4}-\frac{1}{x^{4}}$ is:
Option 1: $8\sqrt2$
Option 2: $18\sqrt2$
Option 3: $6\sqrt2$
Option 4: $24\sqrt2$
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Correct Answer: $24\sqrt2$
Solution :
Given: $x=\sqrt{2}+1$
Thus, $\frac{1}{x}=\sqrt{2}-1$
$x+\frac{1}{x}=(\sqrt{2}+1)+(\sqrt{2}-1)=(\sqrt{2}+1+\sqrt{2}-1)=2\sqrt{2}$
$x-\frac{1}{x}=(\sqrt{2}+1)-(\sqrt{2}-1)=(\sqrt{2}+1-\sqrt{2}+1)=2$
So, $(x^2+\frac{1}{x^2})=6$
We know that,
$x^4-\frac{1}{x^4}=(x^2+\frac{1}{x^2})(x^2-\frac{1}{x^2})=(x^2+\frac{1}{x^2})(x+\frac{1}{x})(x-\frac{1}{x})$
Putting the values we get,
⇒ $x^4-\frac{1}{x^4}=6×2×2\sqrt{2}$
$\therefore x^4-\frac{1}{x^4}=24\sqrt{2}$
Hence, the correct answer is $24\sqrt{2}$.
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