Description for Correct answer:\( \Large x + \frac{1}{x} =3 \) ...(i)
On squaring both sides, we get
\( \Large \left(x + \frac{1}{x}\right)^{2} = \left(3\right)^{2} \)
=> \( \Large x^{2} + \frac{1}{x^{2}} + 2 = 9 \)
=>\( \Large x^{2} + \frac{1}{x^{2}} = 7 \) ...(ii)
Again squaring both sides, we get
\( \Large \left(x^{2} + \frac{1}{x^{2}}\right)^{2} = \left(7\right)^{2} \)
\( \Large x^{4} + \frac{1}{x^{4}} + 2 = 49 \)
\( \Large x^{4} + \frac{1}{x^{4}} = 47 \) ...(iii)
On cubing both sides, we get
\( \Large \left(x + \frac{1}{x}\right)^{3} = \left(3\right)^{3} \)
\( \Large x^{3} + \frac{1}{x^{3}}+ 3\left(x + \frac{1}{x}\right) = 27 \)
\( \Large x^{3} + \frac{1}{x^{3}} + 9 = 27 \) [Because, \( \Large \left(x + \frac{1}{x}\right)=3 \) ]
\( \Large x^{3} + \frac{1}{x^{3}} = 18 \) ...(iv)
On multiplying Eqs. (i) and (iii), we get
\( \Large \left(x^{4} + \frac{1}{x^{4}}\right) \left(x + \frac{1}{x}\right) = 47 \times 3 \)
\( \Large x^{5} + \frac{1}{x^{5}} +x^{3} + \frac{1}{x^{3}} = 141 \)
\( \Large x^{5} + \frac{1}{x^{5}} + 18 = 141 \) [from Eq. (iv)]
\( \Large x^{5} + \frac{1}{x^{5}} = 123 \)
