| \[x+y=4\]and \[\frac{1}{x}+\frac{1}{y}=4,\] then the value of \[{{x}^{3}}+{{y}^{3}}\] is [SSC (CGL) Pre 2014] |
A) 52
B) 64
C) 4
D) 25
Correct Answer: A
Solution :
| Given, \[x+y=4\] ... (i) |
| and \[\frac{1}{x}+\frac{1}{y}=4\] |
| \[\Rightarrow \] \[\frac{y+x}{xy}=4\]\[\Rightarrow \]\[x+y=4xy\] |
| \[\Rightarrow \] \[4=4xy\] [from Eq. (i)] |
| \[\Rightarrow \] \[xy=1\] (ii) |
| We know that, \[{{x}^{3}}+{{y}^{3}}=(x+y)({{x}^{2}}-xy+{{y}^{2}})\] |
| \[{{x}^{3}}+{{y}^{3}}=(x+y)[({{x}^{2}}+{{y}^{2}})-xy]\] |
| \[=(x+y)[{{(x+y)}^{2}}-2xy-xy]\] |
| \[=(x+y)[{{(x+y)}^{2}}-3xy]\] |
| \[=(4)[{{(4)}^{2}}-3\times 1]\] |
| \[=4\,(16-3)=4\times 13=52\] |
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