question_answer

If the roots of the equation \[\left( {{c}^{2}}-ab \right){{x}^{2}}-2\left( {{a}^{2}}-bc \right)x+{{b}^{2}}-ac=0\] in \[x\] are equal, then show that either a\[=0\] or\[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc\].

Answer:

\[\left( {{c}^{2}}-ab \right){{x}^{2}}-2\left( {{a}^{2}}-bc \right)x+{{b}^{2}}-ac=0\]

\[D=0\]

\[\Rightarrow \]                           \[{{b}^{2}}-4ac=0\]

\[\Rightarrow \] \[{{[-2({{a}^{2}}-bc)]}^{2}}-4({{c}^{2}}-ab)({{b}^{2}}-ac)=0\]

\[\Rightarrow \] \[4[{{a}^{4}}+{{b}^{2}}{{c}^{2}}-2{{a}^{2}}bc]-4[{{b}^{2}}{{c}^{2}}-a{{c}^{3}}-a{{b}^{3}}+{{a}^{2}}bc]=0\]

\[\Rightarrow \]     \[4[{{a}^{4}}+{{b}^{2}}{{c}^{2}}-2{{a}^{2}}bc-{{b}^{2}}{{c}^{2}}+a{{c}^{3}}+a{{b}^{3}}-{{a}^{2}}bc]=0\]

\[\Rightarrow \]                                   \[4a[{{a}^{3}}-3abc+{{c}^{3}}+{{b}^{3}}]=0\]

Either \[4a=0\] or \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=0\]

\[a=0\] or \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc\].                                              Hence Proved