| \[{{x}^{2}}+4x+3=0,\] then the value of \[\frac{{{x}^{3}}}{{{x}^{6}}+27{{x}^{3}}+27}\] is |
A) \[-1\]
B) \[-\frac{1}{2}\]
C) 1
D) \[\frac{1}{2}\]
Correct Answer: A
Solution :
| \[{{x}^{2}}+4x+3=0\] |
| \[\Rightarrow \] \[{{x}^{2}}+3x+x+3=0\] |
| \[\Rightarrow \] \[x\,(x+3)+1\,(x+3)=0\] |
| \[\Rightarrow \] \[x=-\,3\] and \[-1\] |
| Now, \[\frac{{{x}^{3}}}{{{x}^{6}}+27{{x}^{3}}+27}\,\,=\,\,\frac{{{x}^{3}}}{{{x}^{3}}\left( {{x}^{3}}+27+\frac{27}{{{x}^{3}}} \right)}\] |
| \[=\,\,\frac{1}{{{x}^{3}}+27+\frac{27}{{{x}^{3}}}}\] |
| On putting \[x=-\,3\] we get |
| \[\frac{1}{{{(-\,3)}^{3}}+27+\frac{27}{{{(-\,3)}^{3}}}}=-1\] |
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