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Question

Answers

A.6

B.9

C.3

D.12

Answer

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For the given system to work, \[{a^3} = {b^3} + {c^3} + {d^3}\] is possible for consecutive natural numbers only if \[a > b > c > d\].

So, let us substitute the values of \[b,c,d\] in form of \[a\], and we get:

\[b = a - 1\]

\[c = a - 2\]

\[d = a - 3\]

Hence, we have \[{a^3} = {\left( {a - 1} \right)^3} + {\left( {a - 2} \right)^3} + {\left( {a - 3} \right)^3}\]

Now, \[a\] cannot be smaller than \[4\] because if it is, then \[d = a - 3\] is going to be less than \[1\] but a natural number cannot be less than \[1\].

Hence, we have established the fact that \[a \ge 4\].

So, let us put in \[a = 4\] and check for the question,

\[\begin{array}{l}LHS = {4^3}\\{\rm{ }} = 64\end{array}\]

\[\begin{array}{l}RHS = {\left( {4 - 1} \right)^3} + {\left( {4 - 2} \right)^3} + {\left( {4 - 3} \right)^3}\\{\rm{ }} = {\left( 3 \right)^3} + {\left( 2 \right)^3} + {\left( 1 \right)^3}\\{\rm{ }} = 27 + 8 + 1\\{\rm{ }} = 36 \ne LHS\end{array}\]

Now, let us try for \[a = 5\],

\[\begin{array}{l}LHS = {5^3}\\{\rm{ }} = 125\end{array}\]

\[\begin{array}{l}RHS = {4^3} + {3^3} + {2^3}\\{\rm{ }} = 64 + 27 + 8\\{\rm{ }} = 99 \ne LHS\end{array}\]

Now, let us try for \[a = 6\],

\[\begin{array}{l}LHS = {6^3}\\{\rm{ }} = 216\end{array}\]

\[\begin{array}{l}RHS = {5^3} + {4^3} + {3^3}\\{\rm{ }} = 125 + 64 + 27\\{\rm{ }} = 216 = LHS\end{array}\]

Hence, the minimum value of \[a = 6\].

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