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Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
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Condition 1)
Since we have a = 1, b = 1, and c = 2, we have
2(a^4 + b^4 + c^4) = 2(1^4 + 1^4 + 2^4)
= 2(1 + 1 + 16)
= 2*18
= 36.
2(a^4 + b4 + c^4) = 36 is a perfect square and the answer is ‘yes’.
Since condition 1) yields a unique solution, it is sufficient.
Condition 2)
Since (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca),
we have a^2 + b^2 + c^2 = -2(ab + bc + ca).
Then, rearranging the second formula gives us:
(a^2 + b^2 + c^2)^2 = (-2(ab + bc + ca))^2
= 4(ab + bc + ca)(ab + bc + ca)
= 4(a^2b^2 + ab^2c + a^2bc + ab^2c + b^2c^2 + abc^2 + a^2bc + abc^2 + a^2c^2)
= 4((a^2b^2 + b^2c^2 + a^2c^2 + 2ab^2c + 2a^2bc + 2abc^2)
= 4(a^2b^2 + b^2c^2 + c^2a^2 + 2abc(a + b + c))
= 4(a^2b^2 + b^2c^2 + c^2a^2), since a + b + c = 0
Following the pattern in the first equation gives us:
(a^2 + b^2 + c^2)^2 = a^4 + b^4 + c^4 + 2(a^2b^2 + b^2c^2 + c^2a^2)
We now have two equations:
(a^2 + b^2 + c^2)^2 = a^4 + b^4 + c^4 + 2(a^2b^2 +b^2c^2+ c^2a^2)
(a^2 + b^2 + c^2)^2 = 4(a^2b^2 + b^2c^2 + c^2a^2)
Combining the two equations gives us:
a^4 + b^4 + c^4 + 2(a^2b^2 + b^2c^2+ c^2a^2) = 4(a^2b^2 + b^2c^2 + c^2a^2)
a^4 + b^4 + c^4 = 2(a^2b^2 + b^2c^2 + c^2a^2)
2(a^4 + b^4 + c^4) = 4(a^2b^2 + b^2c^2 + c^2a^2) = (a^2 + b^2 + c^2)^2.
Thus, 2(a^4 + b^4 + c^4) is a perfect square.
Since condition 2) yields a unique solution, it is sufficient.
Therefore, D is the answer.
Answer: D
This question is a CMT 4(B) question: condition 1) is easy to work with, and condition 2) is difficult to work with. For CMT 4(B) questions, D is most likely the answer.
