If x^2+1/25x^2 =8 3/5 find the value of x^3+1/125x^3 [ ICSE Class 9 S.Chand (Fundamentals of Mathematics) Chapter 3 Expansion and Factorisation Exercise-3.1 Solution ]
$x^2 + \frac{1}{25x^2}$ = $8\frac{3}{5}$
$\Rightarrow \left(x\right)^2 + \left(\frac{1}{5x}\right)^2$ = $\frac{43}{5}$
$\Rightarrow \left(x + \frac{1}{5x}\right)^2 – 2.x\ldotp \frac{1}{5x}$ = $\frac{43}{5}$
$\Rightarrow \left(x + \frac{1}{5x}\right)^2 – \frac{2}{5}$ = $\frac{43}{5}$
$\Rightarrow \left(x + \frac{1}{5x}\right)^2$ = $\frac{43}{5} + \frac{2}{5}$
$\Rightarrow \left(x + \frac{1}{5x}\right)^2$ =$ \frac{45}{5}$
$\Rightarrow \left(x + \frac{1}{5x}\right)^2$ = 9
$\Rightarrow \left(x + \frac{1}{5x}\right)$ = $\pm 3$
$\therefore {\mathrm{x}}^3 + \frac{1}{125{\mathrm{x}}^3}$
= $\left({\mathrm{x}}\right)^3 + \left(\frac{1}{5{\mathrm{x}}}\right)^3$
= $\left({\mathrm{x}} + \frac{1}{5{\mathrm{x}}}\right)^3 – 3.{\mathrm{x}}\ldotp \frac{1}{5{\mathrm{x}}}\left({\mathrm{x}} + \frac{1}{5{\mathrm{x}}}\right)$
Taking , $\left(x + \frac{1}{x}\right)$ = 3
= (3)3 – $\frac{3}{5}$ (3) []
= 27-$\frac{9}{5}$
= $25\frac{1}{5}$
or
Taking, $\left(x + \frac{1}{x}\right)$ = -3
=(-3)3-$\frac{3}{5}$ (-3) [When $\left(x + \frac{1}{x}\right)$ = -3]
= -27 + $\frac{9}{5}$
= -$25\frac{1}{5}$
$\therefore {\mathrm{x}}^3 + \frac{1}{125{\mathrm{x}}^3}$ = $25\frac{1}{5}$ or -$25\frac{1}{5}$