A) 1
B) 64
C) 8
D) None of these
Correct Answer: B
Solution :
Since \[\alpha ,\beta \] are the roots of the equation \[2{{x}^{2}}-35x+2=0\]. Also \[\alpha \beta =1\] \[\therefore \,\,\,2{{\alpha }^{2}}-35\alpha =-2\]or \[2\alpha -35=\frac{-2}{\alpha }\] \[2{{\beta }^{2}}-35\beta =-2\]or \[2\beta -35=\frac{-2}{\beta }\] Now \[{{(2\alpha -35)}^{3}}{{(2\beta -35)}^{3}}={{\left( \frac{-2}{\alpha } \right)}^{3}}{{\left( \frac{-2}{\beta } \right)}^{3}}\] \[=\frac{8.8}{{{\alpha }^{3}}{{\beta }^{3}}}=\frac{64}{1}=64\]
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