| If \[{{x}^{3}}+{{y}^{3}}=9\]and \[x+y=3,\] then the value of \[{{x}^{4}}+{{y}^{4}}\] [SSC (CGL) 2014] |
A) 81
B) 32
C) 27
D) 17
Correct Answer: D
Solution :
| If\[{{x}^{3}}+{{y}^{3}}=9\]and \[x+y=3,\] |
| then \[{{x}^{4}}+{{y}^{4}}=?\] |
| Taking expression \[x+y=3\] |
| On cubing both sides, we get |
| \[{{(x+y)}^{3}}={{(3)}^{3}}\] |
| \[\Rightarrow \]\[{{x}^{3}}+{{y}^{3}}+3xy\,\,(x+y)=27\] |
| \[\Rightarrow \]\[9+3xy\,\,(3)=27\]\[\{\because {{x}^{3}}+{{y}^{3}}=9\,\,\text{and}\,\,x+y=3\}\] |
| \[\Rightarrow \] \[9xy=18\] |
| \[\therefore \] \[xy=2\] |
| Now, \[(x+y)=3\] |
| On squaring both sides, we get |
| \[{{x}^{2}}+{{y}^{2}}+2xy=9\] |
| \[\Rightarrow \]\[{{x}^{2}}+{{y}^{2}}+2\,\,(2)=9\] |
| \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}=5\] \[[\because xy=2]\] |
| Again, squaring on both sides, we get |
| \[{{({{x}^{2}}+{{y}^{2}})}^{2}}={{(5)}^{2}}\] |
| \[\Rightarrow \]\[{{x}^{4}}+{{y}^{4}}+2{{x}^{2}}{{y}^{2}}=25\] |
| \[\Rightarrow \] \[{{x}^{4}}+{{y}^{4}}=25-2\,\,{{(xy)}^{2}}\] |
| \[=25-2\,\,{{(2)}^{2}}=25-8=17\] |
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