Step-by-step explanation:
[tex] \bf➤ \underline{Given-} \\ [/tex]
[tex]\sf{x^{2013} + \frac{1}{x^{2013}} = 2}\\[/tex]
[tex] \bf➤ \underline{To\: find-} \\ [/tex]
[tex]\sf {the\: value \: of \: x^{2022} + \frac{1}{x^{2022}}= ?}\\[/tex]
[tex] \bf ➤\underline{Solution-} \\ [/tex]
Let us assume that:
[tex] \rm: \longmapsto u = {x}^{2013} [/tex]
Therefore, the equation becomes:
[tex] \rm: \longmapsto u + \dfrac{1}{u} = 2[/tex]
[tex] \rm: \longmapsto \dfrac{ {u}^{2} + 1}{u} = 2[/tex]
[tex] \rm: \longmapsto{u}^{2} + 1 = 2u[/tex]
[tex] \rm: \longmapsto{u}^{2} - 2u + 1 =0[/tex]
[tex] \rm: \longmapsto {(u - 1)}^{2} =0[/tex]
[tex] \rm: \longmapsto u = 1[/tex]
Now substitute the value of u. We get:
[tex] \rm: \longmapsto {x}^{2013} = 1[/tex]
[tex] \rm: \longmapsto x = 1[/tex]
Therefore:
[tex] \rm: \longmapsto {x}^{2022} + \dfrac{1}{ {x}^{2022} } = 1 + 1[/tex]
[tex] \rm: \longmapsto {x}^{2022} + \dfrac{1}{ {x}^{2022} } = 2[/tex]
★ Which is our required answer.
[tex]\textsf{\large{\underline{More To Know}:}}[/tex]
(a + b)² = a² + 2ab + b²
(a - b)² = a² - 2ab + b²
a² - b² = (a + b)(a - b)
(a + b)³ = a³ + 3ab(a + b) + b³
(a - b)³ = a³ - 3ab(a - b) - b³
a³ + b³ = (a + b)(a² - ab + b²)
a³ - b³ = (a - b)(a² + ab + b²)
(x + a)(x + b) = x² + (a + b)x + ab
(x + a)(x - b) = x² + (a - b)x - ab
(x - a)(x + b) = x² - (a - b)x - ab
(x - a)(x - b) = x² - (a + b)x + ab
