Do \(P\left( x \right)\) bậc 3 và có 3 nghiệm phân biệt \({x_1},{x_2},{x_3}\) nên \(P\left( x \right)\) được biểu diễn dưới dạng \(P\left( x \right) = a\left( {x - {x_1}} \right)\left( {x - {x_2}} \right)\left( {x - {x_3}} \right){\mkern 1mu} {\mkern 1mu} \left( {a \ne 0} \right)\).

Ta có: \(P'\left( x \right) = a\left( {x - {x_2}} \right)\left( {x - {x_3}} \right) + a\left( {x - {x_1}} \right)\left( {x - {x_3}} \right) + a\left( {x - {x_1}} \right)\left( {x - {x_2}} \right)\)

\(\begin{array}{*{20}{l}}{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{P\left( {{x_1}} \right) = a\left( {{x_1} - {x_2}} \right)\left( {{x_1} - {x_3}} \right)}\\{P\left( {{x_2}} \right) = a\left( {{x_2} - {x_1}} \right)\left( {{x_2} - {x_3}} \right)}\\{P\left( {{x_3}} \right) = a\left( {{x_3} - {x_1}} \right)\left( {{x_3} - {x_2}} \right)}\end{array}} \right.}\\{ \Rightarrow \frac{1}{{P'\left( {{x_1}} \right)}} + \frac{1}{{P'\left( {{x_2}} \right)}} + \frac{1}{{P'\left( {{x_3}} \right)}}}\\{ = \frac{1}{{a\left( {{x_1} - {x_2}} \right)\left( {{x_1} - {x_3}} \right)}} + \frac{1}{{a\left( {{x_2} - {x_1}} \right)\left( {{x_2} - {x_3}} \right)}} + \frac{1}{{a\left( {{x_3} - {x_1}} \right)\left( {{x_3} - {x_2}} \right)}}}\\{ = \frac{{ - {x_2} + {x_3} - {x_3} + {x_1} - {x_1} + {x_2}}}{{a\left( {{x_1} - {x_2}} \right)\left( {{x_2} - {x_3}} \right)\left( {{x_3} - {x_1}} \right)}} = 0}\end{array}\)

Đáp án C.