If (x+1/x)=2 then find (x^2+1/x^2) ,(x^3+1/x^3) and (x^4+1/x^4) [S.Chad ICSE Class 9 Chapter 3 Exercise 3.1 Expansion and Factorisation Solution]
(i) ${\mathrm{x}}^2 + \frac{1}{{\mathrm{x}}^2}$
$\left(x + \frac{1}{x}\right)$ = 2
$\therefore \left(x + \frac{1}{x}\right)^2$ = 4
$\Rightarrow x^2 + 2.x\ldotp \frac{1}{x} + \frac{1}{x^2}$ = 4
$\Rightarrow x^2 + \frac{1}{x^2} + 2$ = 4
$\Rightarrow x^2 + \frac{1}{x^2}$ = 4 – 2
$\Rightarrow x^2 + \frac{1}{x^2}$ = 2
$\therefore x^2 + \frac{1}{x^2}$= 2 [Ans]
(ii) ${\mathrm{x}}^3 + \frac{1}{{\mathrm{x}}^3}$
$\left({\mathrm{x}} + \frac{1}{{\mathrm{x}}}\right)$ = 2
$\therefore \left({\mathrm{x}} + \frac{1}{{\mathrm{x}}}\right)^3$ = 23
$\Rightarrow {\mathrm{x}}^3 + \frac{1}{{\mathrm{x}}^3} + 3.{\mathrm{x}}\ldotp \frac{1}{{\mathrm{x}}}({\mathrm{x}} + \frac{1}{{\mathrm{x}}})$ = 8
$\Rightarrow {\mathrm{x}}^3 + \frac{1}{{\mathrm{x}}^3}$ + 3(2) = 8
$\Rightarrow {\mathrm{x}}^3 + \frac{1}{{\mathrm{x}}^3}$ + 6 = 8
$\Rightarrow {\mathrm{x}}^3 + \frac{1}{{\mathrm{x}}^3}$ = 8 – 6
$\Rightarrow {\mathrm{x}}^3 + \frac{1}{{\mathrm{x}}^3}$ = 2
$\therefore {\mathrm{x}}^3 + \frac{1}{{\mathrm{x}}^3}$ = 2 [Ans]
(iii) ${\mathrm{x}}^4 + \frac{1}{{\mathrm{x}}^4}$
${\mathrm{x}}^2 + \frac{1}{{\mathrm{x}}^2}$ = 2
$\Rightarrow \left({\mathrm{x}}^2 + \frac{1}{{\mathrm{x}}^2}\right)^2$ = 22
$\Rightarrow \left({\mathrm{x}}^2\right)^2 + 2.{\mathrm{x}}^2\ldotp \frac{1}{{\mathrm{x}}^2} + \left(\frac{1}{{\mathrm{x}}^2}\right)^2$ = 4
$\Rightarrow {\mathrm{x}}^4 + \frac{1}{{\mathrm{x}}^4} + 2$ = 4
$\Rightarrow {\mathrm{x}}^4 + \frac{1}{{\mathrm{x}}^4}$ = 4 – 2
$\Rightarrow {\mathrm{x}}^4 + \frac{1}{{\mathrm{x}}^4}$ = 2
$\therefore {\mathrm{x}}^4 + \frac{1}{{\mathrm{x}}^4}$ = 2 [Ans]