A) 1
B) \[5{{(2)}^{2}}-7(2)-6=0\]
C) \[20-14-6=0\]
D) none of these
Correct Answer: B
Solution :
\[\frac{{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca}=\_\_\_\_\_\_\_\_\] \[15{{x}^{2}}+x-28={{15}^{2}}-ax-28,\]then \[11x+12y=58\]
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