Simplify (4a^-8 b^2 / (12a^7 b^-4)) raised to the power 0.

Step-by-step solution

1. 1

   Read the outermost operation first. The entire fraction is raised to the power $0$. Order of operations says that exponent applies to the whole bracket, so the internal structure — however elaborate — is irrelevant until we know whether the base is zero.

2. 2

   Apply the zero-exponent rule. For any nonzero quantity $u$,

   $u^{0}=1$

   The rule follows from the quotient law: $u^{0}=u^{n-n}=\dfrac{u^{n}}{u^{n}}=1$, which is exactly why $u=0$ has to be excluded — it would demand $\tfrac00$.

3. 3

   Check the base is nonzero. The base is $\dfrac{4a^{-8}b^{2}}{12a^{7}b^{-4}}$. Negative exponents already require $a\neq0$ and $b\neq0$, and with those restrictions the numerator $4a^{-8}b^{2}$ is a product of nonzero factors, hence nonzero. So the rule applies:

   $\left(\frac{4a^{-8}b^{2}}{12a^{7}b^{-4}}\right)^{0}=1$

4. 4

   Note the work you were spared. Simplifying the base would give $\dfrac{4}{12}a^{-8-7}b^{2-(-4)}=\dfrac{b^{6}}{3a^{15}}$ — correct, but pointless here, since $\left(\dfrac{b^{6}}{3a^{15}}\right)^{0}=1$ all the same. Recognising the outer exponent first turns a multi-step exercise into a one-line answer.

5. 5

   State the answer with its condition.

   $1,\qquad a\neq0,\ b\neq0$

   The answer is the pure number $1$ — not $0$, and not anything containing $a$ or $b$. A quick check: at $a=b=1$ the base is $\tfrac4{12}=\tfrac13$, and $\left(\tfrac13\right)^{0}=1$ ✓.