A) \[x-{{x}^{-1}}\]
B) \[2x\]
C) \[x+{{x}^{-1}}\]
D) 0
Correct Answer: C
Solution :
[c] \[a={{x}^{\frac{1}{3}}}+{{x}^{\frac{1}{3}}}\] Cubing both sides, we get \[{{a}^{3}}=x+\frac{1}{x}+3({{x}^{\frac{1}{3}}}+{{x}^{\frac{1}{3}}})\] \[{{a}^{3}}=x+\frac{1}{x}+3a\] \[{{a}^{3}}-3a=x+{{x}^{-1}}\]You need to login to perform this action.
You will be redirected in
3 sec