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