Key Property: If \(b^x = b^y\), then \(x = y\) as long as \(x \neq 0\), \(x \neq 1\), and \(x \neq -1\).
The provisos (\(x \neq 0\), \(x \neq 1\), and \(x \neq -1\)) play a large role here!

Given: \(x\) is a non-zero integer, and \((x^x)(x^2) = x^6\)
Let's first apply the Product law to the left side to get: \(x^{x+2} = x^6\)

Now let's assume \(x \neq 0\), \(x \neq 1\), and \(x \neq -1\)
Since the bases are equal, the exponents are equal, which means we can write: \(x+2 = 6\)
So, \(x = 4\) is ONE solution (so far)

We must now test the provisos (other than \(x = 0\), since we're told \(x\) is a nonzero integer)

Let's test \(x = 1\) by plugging it into the equation to get: \(1^{1+2} = 1^6\)
Evaluate to get \(1 = 1\). WORKS!
So, \(x = 1\) is another possible solution.

Now let's test \(x = -1\) by plugging it into the equation to get: \((-1)^{(-1)+2} = (-1)^6\)
Evaluate to get \(-1 = 1\). Doesn't work

So, x can equal 4 or 1.

If \(x = 4\), then Quantity B is greater than Quantity A
If \(x = 1\), then Quantity A is greater than Quantity B

Answer: D

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