Question

If $\frac{x^2+1}{x}=2\frac{1}{2}$, find the values of:
i) $x-\frac{1}{x}$
ii) $x^3-\frac{1}{x^3}$

Answer 1

$x+\frac{1}{x}=\frac{5}{2}$

squaring on both sides

$x^2+\frac{1}{x^2}+2=\frac{25}{4}$

$x^2+\frac{1}{x^2}=\frac{17}{4}$

(i) $x^2+\frac{1}{x^2}-2=\frac{17}{4}-2$

${(x-\frac{1}{x})}^2=\frac{9}{4}\Rightarrow x-\frac{1}{x}=\frac{3}{2}$

(ii) ${(x-\frac{1}{x})}^3={\left(\frac{3}{2}\right)}^3$

$x^3-\frac{1}{x^3}-3(x-\frac{1}{x})=\frac{27}{8}$

$x^3-\frac{1}{x^3}=\frac{27}{8}+3\times \frac{3}{2}$

$x^3-\frac{1}{x^3}=\frac{63}{8}$