A) 5
B) 7
C) 9
D) 11
Correct Answer: B
Solution :
\[a-b=3\] \[\Rightarrow \] \[a=b+3\] ? (i) and \[{{a}^{3}}-{{b}^{3}}=117\] \[\Rightarrow \] \[(a-b)({{a}^{2}}+ab+{{b}^{2}})=117\] ? (ii) Divide equation (ii) by (i), we get \[{{a}^{2}}+ab+{{b}^{2}}=\frac{117}{3}=39\] ? (ii) Put the value of 6 in equation (iii), we get \[{{a}^{2}}+a(a-3)+{{(a-3)}^{2}}=39\] \[\Rightarrow \] \[3{{a}^{2}}-9a+9=39\] \[\Rightarrow \] \[{{a}^{2}}-3a-10=0\] \[\Rightarrow \] \[(a+2)(a-5)=0\] \[\therefore \] \[a=-2\]or\[5\]and\[b=-5\]or\[2\] \[\therefore \] \[a+6=5+2=7\]
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