| If \[\frac{a}{b}+\frac{b}{a}=1,\]then the value of \[({{a}^{3}}+{{b}^{3}})\] is [SSC (CGL) 2007] |
A) 0
B) 1
C) 2
D) 3
Correct Answer: A
Solution :
| Given, \[\frac{a}{b}+\frac{b}{a}=1\] |
| \[\Rightarrow \] \[\frac{{{a}^{2}}+{{b}^{2}}}{ab}=1\] |
| \[\Rightarrow \] \[{{a}^{2}}+{{b}^{2}}=ab\] (i) |
| Now, \[{{a}^{3}}+{{b}^{3}}=({{a}^{2}}+{{b}^{2}}-ab)\] |
| \[=(a+b)(ab-ab)\] [from Eq.(i)] |
| \[=(a+b)\cdot 0=0\] |
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