A) \[\Rightarrow \]
B) \[P(-a)=0\]
C) \[P(x)={{x}^{3}}-3{{x}^{2}}+4x-12\]
D) none of these
Correct Answer: C
Solution :
Let y be the additive inverse of \[{{\left( \frac{a+b}{a-b} \right)}^{2}}\]. Then, \[\frac{{{x}^{5}}-7{{x}^{2}}+18}{{{x}^{3}}-2}+Y=0\] \[\Rightarrow \]\[y=\frac{-{{x}^{5}}+7{{x}^{2}}-18}{{{x}^{3}}-2}\]
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