If x_{1} x_{2} x_{3} = 4(4 + x_{1} + x_{2} + | vedclass

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(B) Let $S = \frac{1}{2+x_{1}} + \frac{1}{2+x_{2}} + \frac{1}{2+x_{3}}$.Given $x_{1} x_{2} x_{3} = 16 + 4(x_{1} + x_{2} + x_{3})$.Consider the express

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(B) Let $S = \frac{1}{2+x_{1}} + \frac{1}{2+x_{2}} + \frac{1}{2+x_{3}}$.Given $x_{1} x_{2} x_{3} = 16 + 4(x_{1} + x_{2} + x_{3})$.Consider the expression $P = (2+x_{1})(2+x_{2})(2+x_{3}) = 8 + 4(x_{1} + x_{2} + x_{3}) + 2(x_{1}x_{2} + x_{2}x_{3} + x_{3}x_{1}) + x_{1}x_{2}x_{3}$.Substituting $x_{1}x_{2}x_{3} = 16 + 4(x_{1} + x_{2} + x_{3})$,we get $P = 8 + 4(x_{1} + x_{2} + x_{3}) + 2(x_{1}x_{2} + x_{2}x_{3} + x_{3}x_{1}) + 16 + 4(x_{1} + x_{2} + x_{3}) = 24 + 8(x_{1} + x_{2} + x_{3}) + 2(x_{1}x_{2} + x_{2}x_{3} + x_{3}x_{1})$.The numerator of the sum $S$ is $(2+x_{2})(2+x_{3}) + (2+x_{1})(2+x_{3}) + (2+x_{1})(2+x_{2}) = 12 + 4(x_{1} + x_{2} + x_{3}) + (x_{1}x_{2} + x_{2}x_{3} + x_{3}x_{1})$.Using the relation $x_{1}x_{2}x_{3} = 16 + 4(x_{1} + x_{2} + x_{3})$,we can simplify the expression to show that $S = \frac{1}{2}$.

If $x_{1} x_{2} x_{3} = 4(4 + x_{1} + x_{2} + x_{3})$,then what is the value of $\left[\frac{1}{(2 + x_{1})}\right] + \left[\frac{1}{(2 + x_{2})}\right] + \left[\frac{1}{(2 + x_{3})}\right]$?

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