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Question Number 164549 by wwww last updated on 18/Jan/22

Answered by alephzero last updated on 18/Jan/22

$$\left(a\right)a-\frac{1}{a}=4$$$$\frac{a^2-1}{a}=4$$$$\frac{a^2-1-4a}{a}=0$$$$\Rightarrow a^2-4a-1=0$$$$a=\frac{4\pm \sqrt{16+4}}{2}=\frac{4\pm \sqrt{20}}{2}=\frac{4\pm 2\sqrt{5}}{2}=$$$$=2\pm \sqrt{5}$$$$\mathrm{Let}a+\frac{1}{a}=\mathcal{A}$$$$\Rightarrow {\mathcal{A}}_1=2+\sqrt{5}+\frac{1}{2+\sqrt{5}}=2+\sqrt{5}-2+\sqrt{5}=$$$$=2\sqrt{5}$$$${\mathcal{A}}_2=2-\sqrt{5}+\frac{1}{2-\sqrt{5}}=2-\sqrt{5}-2-\sqrt{5}=$$$$=-2\sqrt{5}$$$$\Rightarrow a+\frac{1}{a}=\pm 2\sqrt{5}$$$$\left(b\right)a^3+\frac{1}{a^3}=\mathcal{B}$$$$a_1^3=38+17\sqrt{5}\wedge a_2^3=38-17\sqrt{5}$$$$\Rightarrow {\mathcal{B}}_1=38+17\sqrt{5}+\frac{1}{38+17\sqrt{5}}=34\sqrt{5}$$$${\mathcal{B}}_2=38-17\sqrt{5}+\frac{1}{38-17\sqrt{5}}=-34\sqrt{5}$$

Answered by Rasheed.Sindhi last updated on 18/Jan/22

$$a-\frac{1}{a}=4$$$$\left(a\right):{(a-\frac{1}{a}=4)}^2$$$$a^2+\frac{1}{a^2}-2=16$$$${(a+\frac{1}{a})}^2=20$$$$a+\frac{1}{a}=\pm 2\sqrt{5}$$$$\left(b\right):{(a+\frac{1}{a})}^3={(\pm 2\sqrt{5})}^3$$$$a^3+\frac{1}{a^3}+3(a+\frac{1}{a})=\pm 40\sqrt{5}$$$$a^3+\frac{1}{a^3}+3(\pm 2\sqrt{5})=\pm 40\sqrt{5}$$$$a^3+\frac{1}{a^3}=\pm 40\sqrt{5}\mp 6\sqrt{5}=\pm 34\sqrt{5}$$$$\left(c\right):{(a-\frac{1}{a}=4)}^3$$$$a^3-\frac{1}{a^3}-3(a-\frac{1}{a})=64$$$$a^3-\frac{1}{a^3}-3\left(4\right)=64$$$$a^3-\frac{1}{a^3}=76$$$$(a^3-\frac{1}{a^3})(a^3+\frac{1}{a^3})=\left(76\right)(\pm 34\sqrt{5})$$$$a^6-\frac{1}{a^6}=\pm 2584\sqrt{5}$$$$$$

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