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The real value of a for which the roots $x_1,x_2,x_3$ of $x^3-6x^2+ax-a=0$ satisfy $(x_1-3)^3+(x_2-3)^3+(x_3-3)^3=0$ is

$\begin{array}{1 1}(A)\;3&(B)\;-3\\(C)\;9&(D)\;-9\end{array} $

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$x_1+x_2+x_3=6$
$(x_1+x_2+x_3)^2-2(x_1x_2+x_2x_3+x_3x_4)=36-2a$
$\Rightarrow x_1^2+x_2^2+x_3^2=36-2a$
$x_1^3+x_2^3+x_3^3=6(x_1^2+x_2^2+x_3^2)-a(x_1+x_2+x_3)+3a$
$\Rightarrow 216-21a$
$(x_1-3)^3+(x_2-3)^3+(x_3-3)^3=0$
$\Rightarrow x_1^3+x_2^3+x_3^3-9(x_1^2+x_2^2+x_3^2)+27(x_1+x_2+x_3)-243=0$
$\Rightarrow -27-3a=0$
$\Rightarrow a=-9$
Hence (D) is the correct answer.
answered Apr 16, 2014 by sreemathi.v
 

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