Key property: Any non-zero number raised to an EVEN power will evaluate to be a POSITIVE number.

Given: \((az) ^{79} \times (bz) ^{71} \times (cz) ^8 \times (dz) ^2 > 0\)

Rewrite as: \(a ^{79} \times z^{79} \times b^{71} \times z^{71} \times c^8 \times z^8 \times d^2 \times z^2 > 0\)

Combine all of the \(z\) terms: \(a ^{79} \times b^{71} \times c^8 \times d^2 \times z^{160} > 0\)

Since we can be certain that \(c^8\) is positive, we can safely divide both sides of the inequality by \(c^8\) to get: \(a ^{79} \times b^{71} \times d^2 \times z^{160} > 0\)

For the same reason, we can safely divide both sides of the inequality by \(d^2\) to get: \(a ^{79} \times b^{71} \times z^{160} > 0\)

And, for the same reason, we can safely divide both sides of the inequality by \(z^{160}\) to get: \(a ^{79} \times b^{71} > 0\)

Since we can be certain that \(a^{78}\) is positive, we can safely divide both sides of the inequality by \(a^{78}\) to get: \(a ^{1} \times b^{71} > 0\)

Since we can be certain that \(b^{70}\) is positive, we can safely divide both sides of the inequality by \(b^{70}\) to get: \(a ^{1} \times b^{1} > 0\)

In other words, \(ab > 0\)

Answer: B