A) 1
B) 0
C) 3
D) -1
Correct Answer: B
Solution :
(b): We have, \[P=2-a\Rightarrow a+p-2=0\] Now, \[{{a}^{3}}+6ap+{{p}^{3}}-8={{a}^{3}}+{{p}^{3}}+{{\left( -2 \right)}^{3}}-3ap(-2)\] \[=\left\{ a+p+\left( -2 \right) \right\}\left\{ {{a}^{2}}+{{p}^{2}}+{{(-2)}^{2}}-ap-p\left( -2 \right)-a\left( -2 \right) \right\}\] \[=\left\{ a+p-2 \right\}\left\{ {{a}^{2}}+{{p}^{2}}+4-ap+2p+2a \right\}\] \[=0\times \left( {{a}^{2}}+{{p}^{2}}+4-ap+2p+2a \right)=0.\]You need to login to perform this action.
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