If \[{{(a+b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\], then the value of \[x+\frac{1}{1+\frac{1}{1+\frac{1}{x}}}\]is equal to

• A.  \[{{(a+b)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}={{a}^{3}}+{{b}^{3}}+3ab(a+b)\] 
• B.  \[{{(a-b)}^{3}}={{a}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}-{{b}^{3}}={{a}^{3}}-{{b}^{3}}-3ab(a-b)\]
• C.  \[{{a}^{3}}+{{b}^{3}}=(a+b)({{a}^{2}}-ab+{{b}^{2}})\]         
• D. none of these