Question

If$f\left(x\right)=x+\frac{1}{x},$ prove that $\left[f\right(x){]}^3=f(x{)}^3+3f\left(\frac{1}{x}\right)$. Also determine the value of $\left[f\right(2){]}^3.$

Answer 1

We have $f\left(x\right)=x+\frac{1}{x}$

$RHS=f\left(x^3\right)+3f\left(\frac{1}{x}\right)=x^3+\frac{1}{x^3}+3(\frac{1}{x}+x)$

$=x^3+3x+\frac{3}{x}+\frac{1}{x^3}$

$={(x+\frac{1}{x})}^3[\because (a+b{)}^3=a^3+b3+3a^{2}b+ba{b}^2]$

$=\left[f\right(x){]}^3=RHS$

$\therefore \left[f\right(x){]}^3+3f\left(\frac{1}{x}\right)Henceproved...(i)$

$putx=2inEq.\left(i\right)weget$

${\left[f\right(2\left)\right]}^3=f(2{)}^3+3f\left(\frac{1}{2}\right)(i)$

$Now,f\left(2^3\right)=f\left(8\right)=8+\frac{1}{8}=\frac{65}{8}andf\left(\frac{1}{2}\right)=\frac{1}{2}+\frac{1}{1}=\frac{1}{2}+2=\frac{5}{2}$

from E.g (ii), we get

${\left[f\right(2\left)\right]}^3=\frac{65}{8}+3\times \frac{5}{2}=\frac{65}{8}+\frac{15}{2}=\frac{65+60}{8}=\frac{125}{2}$