If a, b, c and d are consecutive even integers such that a<b<c<d, whic

\(a^2+(a+2)^2+(a+4)^2-(a+6)^2\)

\(a^2+a^2+4a+4+a^2+8a+16-a^2-12a-36\)

\(2a^2-16\)

Let's check the answer choices.

A. \(2a^2-16 = -24\) would mean \(2a^2\) would be negative, but it can't be. Eliminate.

B. \(2a^2-16 = -16\) means \(2a^2 = 0\), which means \(a^2 = 0\), which means a=0. That is allowed.

Answer choice B.

Since a, b, c, and d are consecutive even integers, b = a + 2, c = a + 4, and d = a + 6. Let's substitute these expressions for b, c, and d in \(a^2 + b^2 + c^2 - d^2\) and simplify:

\(\Rightarrow a^2 + b^2 + c^2 - d^2\)

\(\Rightarrow a^2 + (a + 2)^2 + (a + 4)^2 - (a + 6)^2\)

\(\Rightarrow a^2 + (a^2 + 4a + 4) + (a^2 + 8a + 16) - (a^2 + 12a + 36)\)

\(\Rightarrow 3a^2 + 12a + 20 - a^2 - 12a - 36\)

\(\Rightarrow 2a^2 - 16\)

It is now easy to see that if a = 0, then \(2a^2 - 16 = -16\), which means that -16 is a possible value for \(a^2 + b^2 + c^2 - d^2\).

Alternatively, we can set each answer choice equal to \(2a^2 - 16\) and solve:

\(\bullet\) If \(2a^2 - 16 = -24\), then \(2a^2 = -8\), and \(a^2 = -4\), which is not possible.
\(\bullet\) If \(2a^2 - 16 = -16\), then \(2a^2 = 0\), which means a = 0. Since 0 is an even integer, -16 is a possible value of \(a^2 + b^2 + c^2 - d^2\).
\(\bullet\) If \(2a^2 - 16 = 8\), then \(2a^2 = 24\). Simplifying, we get \(a^2 = 12\), which means \(a = \pm 2\sqrt{3}\). Since neither \(2\sqrt{3}\) nor \(-2\sqrt{3}\)is not an even integer, 8 is not a possible value of \(a^2 + b^2 + c^2 - d^2\).
\(\bullet\) If \(2a^2 - 16 = 48\), then \(2a^2 = 64\). Simplifying, we get \(a^2 = 32\), which means \(a = \pm 4\sqrt{2}\). Since neither \(4\sqrt{2}\) nor \(-4\sqrt{2}\) is not an even integer, 48 is not a possible value of \(a^2 + b^2 + c^2 - d^2\).
\(\bullet\) If \(2a^2 - 16 = 88\), then \(2a^2 = 104\). Simplifying, we get \(a^2 = 52\), which means \(a = \pm 2\sqrt{13}\). Since neither \(2\sqrt{13}\) nor \(-2\sqrt{13}\)is not an even integer, 88 is not a possible value of \(a^2 + b^2 + c^2 - d^2\).

Answer: B­