If x^2+1/x^2=18 find the value of (x-1/x) and (x^3-1/x^3) [S. Chand Class 9 Math Solution Chapter 3 Exercise 3.1 Expansion and Factorisation ]
$\left(x – \frac{1}{x}\right)^2$ =
$x^2 – 2.x\ldotp \frac{1}{x} + \frac{1}{x^2}$
= $x^2 + \frac{1}{x^2} – 2$
= 18 – 2
= 16
$\therefore \left({\mathrm{x}} – \frac{1}{{\mathrm{x}}}\right)$ = $\pm 4$
Now, $\left({\mathrm{x}} – \frac{1}{{\mathrm{x}}}\right)^3$ = ${\mathrm{x}}^3 – \frac{1}{{\mathrm{x}}^3} – 3.{\mathrm{x}}\ldotp \frac{1}{{\mathrm{x}}}\ldotp \left({\mathrm{x}} – \frac{1}{{\mathrm{x}}}\right)$
$\Rightarrow \left(4\right)^3$ = ${\mathrm{x}}^3 – \frac{1}{{\mathrm{x}}^3} – 3(4)$ [When x-1/x=4]
$\Rightarrow 64$ = ${\mathrm{x}}^3 – \frac{1}{{\mathrm{x}}^3} – 12$
$\Rightarrow {\mathrm{x}}^3 – \frac{1}{{\mathrm{x}}^3}$ = 64 + 12
$\Rightarrow {\mathrm{x}}^3 – \frac{1}{{\mathrm{x}}^3}$ = 76
Again, $\left({\mathrm{x}} – \frac{1}{{\mathrm{x}}}\right)^3$ = ${\mathrm{x}}^3 – \frac{1}{{\mathrm{x}}^3} – 3.{\mathrm{x}}\ldotp \frac{1}{{\mathrm{x}}}\ldotp \left({\mathrm{x}} – \frac{1}{{\mathrm{x}}}\right)$
$\Rightarrow \left(-4\right)^3$ = ${\mathrm{x}}^3 – \frac{1}{{\mathrm{x}}^3} – 3(-4)$ [When x-1/x=-4]
$\Rightarrow -64$ = ${\mathrm{x}}^3 – \frac{1}{{\mathrm{x}}^3} + 12$
$\Rightarrow {\mathrm{x}}^3 – \frac{1}{{\mathrm{x}}^3}$ = -64 -12
$\Rightarrow {\mathrm{x}}^3 – \frac{1}{{\mathrm{x}}^3}$ = -76
$\therefore {\mathrm{x}}^3 – \frac{1}{{\mathrm{x}}^3}$ = $\pm 76$