A) 194
B) 196
C) 198
D) 200
Correct Answer: C
Solution :
\[-bc,bd\in I,bd\ne 0\] Similarly, \[b=3-2\sqrt{2}\] We have, \[{{(a+b)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\,(a+b)\] \[\frac{a}{b}+\frac{c}{d}(c\ne 0)=\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{ad}{bc}\] \[ad,bc\in I,bc\ne 0\]\[\left( 3+2\sqrt{2}+3\text{-}2\sqrt{2} \right)\] \[p,q,r,s,l,m\in I,q\ne 0,s\ne 0,m\ne 0.\]
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