| If \[x+y=9\] and \[\frac{1}{x}+\frac{1}{y}=3,\] then the value of \[{{x}^{3}}+{{y}^{3}}\] is |
A) 645
B) 459
C) 729
D) 648
Correct Answer: D
Solution :
| \[x+y=9\] and \[\frac{1}{x}+\frac{1}{y}=3\] |
| \[\Rightarrow \] \[\frac{x+y}{xy}=3\]\[\Rightarrow \]\[xy=\frac{x+y}{3}=\frac{9}{3}=3.\] |
| Now, \[({{x}^{3}}+{{y}^{3}})=(x+y)({{x}^{2}}+{{y}^{2}}-xy)\] |
| \[=(x+y)[{{(x+y)}^{2}}-2xy-xy]\] |
| \[=(x+y)[{{(x+y)}^{2}}-3xy]\] |
| \[=9\,[{{9}^{2}}-3\times 3]=9\,[81-9]\] |
| \[=9\times 72=648\] |
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