A) \[{{a}^{3}}+{{b}^{3}}\]
B) \[{{(a-b)}^{3}}\]
C) \[{{a}^{3}}-{{b}^{3}}\]
D) None of these
Correct Answer: B
Solution :
If one root is square of other of the equation\[a{{x}^{2}}+bx+c=0\], then \[{{b}^{3}}+a{{c}^{2}}+{{a}^{2}}c=3abc\] Which can be written in the form \[a{{(c-b)}^{3}}=c{{(a-b)}^{3}}\] Trick: Let roots be 2 and 4, then the equation is\[{{x}^{2}}-6x+8=0\]. Here obviously \[X=\frac{a{{(c-b)}^{3}}}{c}=\frac{1{{(14)}^{3}}}{8}=\frac{14}{2}\times \frac{14}{2}\times \frac{14}{2}={{7}^{3}}\] Which is given by\[{{(a-b)}^{3}}={{7}^{3}}\].
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