Discussion

Explanation:

Given, x⁸ + ${\left(\frac{1}{\mathrm{x}}\right)}^8$ = 47

⇒ x⁸ + ${\left(\frac{1}{\mathrm{x}}\right)}^8$ + 2 = 49

⇒ (x⁴)² + ${\left(\frac{1}{{\mathrm{x}}^4}\right)}^2$ + $2\times {\mathrm{x}}^4\times \frac{1}{{\mathrm{x}}^4}$ = 49

⇒ ${\left({\mathrm{x}}^4+\frac{1}{{\mathrm{x}}^4}\right)}^2$ = 49

⇒ ${\mathrm{x}}^4+\frac{1}{{\mathrm{x}}^4}$ = 7 [-7 will be rejected as LHS should be positive]

Now, ${\mathrm{x}}^4+\frac{1}{{\mathrm{x}}^4}$ + 2 = 9

⇒ ${\left({\mathrm{x}}^2\right)}^2+{\left(\frac{1}{{\mathrm{x}}^2}\right)}^2$ + $2\times {\mathrm{x}}^2\times \frac{1}{{\mathrm{x}}^2}$ = 9

⇒ ${\left({\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2}\right)}^2$ = 9

⇒ ${\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2}$ = 3

Similarly, $\mathrm{x}+\frac{1}{\mathrm{x}}$ = $\sqrt{5}$

Cubing both sides, we get

⇒ x³ + $\frac{1}{{\mathrm{x}}^3}$ + $3\times \mathrm{x}\times \frac{1}{\mathrm{x}}\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)$ = 5√5

⇒ x³ + $\frac{1}{{\mathrm{x}}^3}$ = 2√5

Again cubing both sides, we get

⇒ x⁹ + $\frac{1}{{\mathrm{x}}^9}$ + $3\times {\mathrm{x}}^3\times \frac{1}{{\mathrm{x}}^3}\left({\mathrm{x}}^3+\frac{1}{{\mathrm{x}}^3}\right)$ = 40√5

⇒ x⁹ + $\frac{1}{{\mathrm{x}}^9}$ = 34√5

Hence, option (a).

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