A) \[{{(ac)}^{1/3}}+{{(ab)}^{1/3}}+c=0\]
B) \[{{({{a}^{3}}b)}^{1/4}}+{{(a{{b}^{3}})}^{1/4}}+c=0\]
C) \[{{({{a}^{3}}c)}^{1/4}}+{{(a{{c}^{3}})}^{1/4}}+b=0\]
D) \[{{({{a}^{4}}c)}^{1/3}}+{{(a{{c}^{4}})}^{1/3}}+b=0\]
E) \[{{({{a}^{3}}c)}^{1/4}}-{{(a{{c}^{3}})}^{1/4}}+b=0\]
Correct Answer: C
Solution :
\[\because \]\[\alpha \]and\[\beta \]are the roots of equation \[a{{x}^{2}}+bx+c=0.\] \[\alpha +\beta =-\frac{b}{a}\]and \[\alpha \beta =\frac{c}{a}\] \[\Rightarrow \] \[\alpha +{{\alpha }^{1/3}}=-\frac{b}{a}\]and\[{{\alpha }^{4/3}}=\frac{c}{a}\] \[(\because \beta ={{\alpha }^{1/3}})\] \[\Rightarrow \] \[{{\left( \frac{c}{a} \right)}^{3/4}}+{{\left( \frac{c}{a} \right)}^{1/4}}=-\frac{b}{a}\] \[\Rightarrow \] \[{{({{a}^{3}}c)}^{1/4}}+{{(a{{c}^{3}})}^{1/4}}=-\frac{b}{a}\]
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