Question

If $f\left(x\right)=x+\frac{1}{x},x\ne 0;\mathrm{then}{\left(f\left(x\right)\right]}^3=$

• A. $f\left(x^3\right)+3f\left(x\right)$
• B. $f\left(x^3\right)-3f\left(x\right)$
• C. $f\left(x^3\right)+3f\left(\frac{1}{x}\right)$
• D. $f\left(x^3\right)-3f\left(\frac{1}{x}\right)$

Answer 1

Answer: (A), (C)

The correct options are
A $f\left(x^3\right)+3f\left(x\right)$
C $f\left(x^3\right)+3f\left(\frac{1}{x}\right)$

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