A) 104
B) 108
C) 112
D) 5
Correct Answer: B
Solution :
| \[\alpha\] and \[\beta\]are the zeroes of \[f\left( x \right)={{x}^{2}}-4x+3\] |
| We know that, |
| \[\alpha +\beta =-\frac{b}{a}=4\] |
| and \[\alpha \,.\,\beta =\frac{c}{a}=3\] |
| \[\Rightarrow \,\,\,{{\alpha }^{4}}{{\beta }^{3}}+{{\alpha }^{3}}{{\beta }^{4}}={{\alpha }^{3}}{{\beta }^{3}}\left( \alpha +\beta \right)\] |
| \[\Rightarrow \,\,{{\left( \alpha \,\beta \right)}^{3}}\left( \alpha +\beta \right)={{\left( 3 \right)}^{3}}\left( 4 \right)\] |
| \[=4\times 27=108\] |
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