If $$x=\sqrt{2}+1$$, then the value of $$x^4-\frac{1}{x^4}$$ is

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Given : $$x=\sqrt{2}+1$$

=> $$x^2=(\sqrt2+1)^2$$

=> $$x^2=2+1+2\sqrt2$$

=> $$x^2=3+2\sqrt2$$

Squaring both sides, => $$x^4=9+8+2(3)(2\sqrt2)$$

=> $$x^4=17+12\sqrt2$$ ---------------(i)

=> $$\frac{1}{x^4}=\frac{1}{17+12\sqrt2}$$

=> $$\frac{1}{x^4}=\frac{1}{17+12\sqrt2} \times \frac{(17-12\sqrt2)}{(17-12\sqrt2)}$$

=> $$\frac{1}{x^4}=\frac{17-12\sqrt2}{(17)^2-(12\sqrt2)^2}$$

=> $$\frac{1}{x^4}=\frac{17-12\sqrt2}{289-288}$$

=> $$\frac{1}{x^4}=17-12\sqrt2$$ -----------(ii)

Subtracting equation (ii) from (i)

$$\therefore$$ $$x^4-\frac{1}{x^4}=(17+12\sqrt2)-(17-12\sqrt2)$$

= $$12\sqrt2+12\sqrt2=24\sqrt2$$

=> Ans - (D)