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Question Number 90406 by I want to learn more last updated on 23/Apr/20

$$\mathrm{If}\mathrm{x}-\frac{1}{\mathrm{x}}=3$$$${\mathrm{x}}^4-\frac{1}{{\mathrm{x}}^4}=???$$

Answered by som(math1967) last updated on 23/Apr/20

$$x-\frac{1}{x}=3$$$$\Rightarrow {(x-\frac{1}{x})}^2=9$$$$\Rightarrow {(x+\frac{1}{x})}^2-4x.\frac{1}{x}=9$$$$\Rightarrow (x+\frac{1}{x})=\pm \sqrt{13}$$$$again{(x-\frac{1}{x})}^2=9$$$$\therefore x^2+\frac{1}{x^2}=9+2=11$$$$x^4-\frac{1}{x^4}=(x^2+\frac{1}{x^2})(x+\frac{1}{x})(x-\frac{1}{x})$$$$=\pm 33\sqrt{13}ans$$

Commented by I want to learn more last updated on 23/Apr/20

$$\mathrm{I}\mathrm{appreciate}\mathrm{sir}$$

Answered by MJS last updated on 23/Apr/20

$$x=u+v=\frac{{(u+v)}^2(u-v)}{(u+v)(u-v)}$$$$\frac{1}{x}=\frac{u-v}{(u+v)(u-v)}$$$$x^4-\frac{1}{x^4}=\frac{{(u+v)}^8{(u-v)}^4-{(u-v)}^4}{{(u+v)}^4{(u-v)}^4}=$$$$=\frac{{(u^2-v^2)}^4{(u+v)}^4-{(u-v)}^4}{{(u^2-v^2)}^4}$$$$x-\frac{1}{x}=3\Rightarrow x^2-3x-1=0\Rightarrow x=\frac{3}{2}\pm \frac{\sqrt{13}}{2}$$$$u=\frac{3}{2}\wedge v^2=\frac{13}{4}$$$$\frac{{(\frac{9}{4}-\frac{13}{4})}^4{(u+v)}^4-{(u-v)}^4}{{(\frac{9}{4}-\frac{13}{4})}^4}=$$$$={(u+v)}^4-{(u-v)}^4=8uv(u^2+v^2)=$$$$=8uv(\frac{9}{4}+\frac{13}{4})=44uv=\pm 33\sqrt{13}$$

Commented by I want to learn more last updated on 23/Apr/20

$$\mathrm{I}\mathrm{appreciate}\mathrm{sir}$$

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