• # question_answer If one root of the equation $a{{x}^{2}}+bx+c=0$the square of the other, then$a{{(c-b)}^{3}}=cX$, where X is A) ${{a}^{3}}+{{b}^{3}}$ B) ${{(a-b)}^{3}}$ C) ${{a}^{3}}-{{b}^{3}}$ D) None of these

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${{(a-b)}^{3}}={{7}^{3}}$.