A) \[3abc\]
B) \[-3abc\]
C) 0
D) None of these
Correct Answer: A
Solution :
Let\[\alpha ,{{\alpha }^{2}}\]be the two roots. Then \[\alpha +{{\alpha }^{2}}=-\frac{b}{a}\] .....(i) and \[\alpha .{{\alpha }^{2}}=\frac{c}{a}\] .....(ii) On cubing both sides of (i) \[{{\alpha }^{3}}+{{\alpha }^{6}}+3\alpha {{\alpha }^{2}}(\alpha +{{\alpha }^{2}})=-\frac{{{b}^{3}}}{{{a}^{3}}}\] \[\Rightarrow \frac{c}{a}+\frac{{{c}^{2}}}{{{a}^{2}}}+3\frac{c}{a}\left( -\frac{b}{a} \right)=-\frac{{{b}^{3}}}{{{c}^{3}}}\] [By (i) and (ii)] \[\Rightarrow {{b}^{3}}+a{{c}^{2}}+{{a}^{2}}c=3abc\].
You need to login to perform this action.
You will be redirected in
3 sec
