Question

$\left[\frac{(a^0+b^0)(a^0-b^0)}{a^2-b^2}\right]^{a^4}$ (where a ≠ 0 and b ≠ 0) is

• A. 0
• B. 1
• C. -1
• D. Not defined

Answer 1

Answer: (A)

0

Answer 2

Here's the solution:

1. Recognize the zero exponent rule: Any non-zero number raised to the power of 0 is 1. Therefore, $a^0 = 1$ and $b^0 = 1$.

2. Substitute: $\left[\frac{(1+1)(1-1)}{a^2-b^2}\right]^{a^4}$

3. Simplify the numerator: $(1+1)(1-1) = (2)(0) = 0$

4. Simplify the expression: $\left[\frac{0}{a^2-b^2}\right]^{a^4} = [0]^{a^4}$

5. Evaluate: Since $a \ne 0$, $a^4$ will be a positive number. Therefore, $0$ raised to any positive power is $0$.

   $0^{a^4} = 0$

Therefore, the answer is 0.