xả !!!!!!!!

\(A^3=n+1+n-1+3\sqrt[3]{n+1}.\sqrt[3]{n-1}\left(\sqrt[3]{n+1}+\sqrt[3]{n-1}\right)=2n+3\sqrt[3]{\left(n-1\right)\left(n+1\right)}\left(\sqrt[3]{n+1}+\sqrt[3]{n-1}\right)\)Vì n >=1 nên A³ > B³ => A > B

I read the contract again and again _____ avoiding making spelling mistakes.

A. in terms of

B. by means of

C. with a view to

D. in view of

phải ko