A) 5
B) 48
C) 20
D) 22
Correct Answer: C
Solution :
| [c] Given, \[{{a}^{3}}-{{b}^{3}}=56\] (i) \[a-b=2\] (ii) Using Eq. (ii), we get \[a=2+b\] on cubing both sides, we get \[{{a}^{3}}={{(2)}^{3}}+{{(b)}^{3}}+3\times 2\times b(2+b)\] \[\Rightarrow \] \[{{a}^{3}}=8+{{b}^{3}}+6b\,(2+b)\] On putting value of \[{{a}^{3}}\] in Eq. (i), we get \[8+{{b}^{3}}+6b\,(2+b)-{{b}^{3}}=56\] \[\Rightarrow \] \[6b\,(2+b)=56-8\] \[\Rightarrow \] \[12b+6{{b}^{2}}=48\] \[\Rightarrow \] \[6{{b}^{2}}+12b-48=0\] \[\Rightarrow \] \[{{b}^{2}}-2b-8=0\] \[\Rightarrow \] \[{{b}^{2}}+4b-2b-8=0\] \[\Rightarrow \] \[b\,(b+4)-2\,(b-4)=0\] \[\Rightarrow \] \[(b-2)(b+4)=0\] \[\therefore \] \[b=2,\]\[-\,4\] Taking positive value because b is positive. Then, \[{{a}^{3}}-{{b}^{3}}=56\] \[\Rightarrow \] \[{{a}^{3}}=56+{{(2)}^{3}}\] \[\Rightarrow \] \[{{a}^{3}}=56+{{(2)}^{3}}\] \[\Rightarrow \] \[a=\sqrt[3]{64}=4\] Now, \[{{a}^{2}}+{{b}^{2}}={{(4)}^{2}}+{{(2)}^{2}}\] \[=16+4=20\] |
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