Simplify 125\(^{−\frac{1}{3}}\) × 49\(^{−\frac{1}{2}}\) × 10\(^0\)

Answer Details

Start by breaking down each part of the expression:

• \( 125^{-\frac{1}{3}} \)
• \( 49^{-\frac{1}{2}} \)
• \( 10^{0} \)

Let’s look at each part:

• Exponent rules: For any nonzero number \( a \), \( a^0 = 1 \). So, \( 10^0 = 1 \).
• Negative exponents: \( a^{-n} = \frac{1}{a^n} \). Negative exponents mean "take the reciprocal."
• Fractional exponents: \( a^{\frac{1}{n}} = \sqrt[n]{a} \). The denominator in the fraction tells what root to take.

Let's simplify each part:

• \( 125^{-\frac{1}{3}} \) means take the reciprocal and the cube root:
  \[ 125^{-\frac{1}{3}} = \frac{1}{125^{1/3}} \] The cube root of 125 is 5 (because \( 5^3 = 125 \)), so: \[ 125^{-\frac{1}{3}} = \frac{1}{5} \]
• \( 49^{-\frac{1}{2}} \) means take the reciprocal and the square root:
  \[ 49^{-\frac{1}{2}} = \frac{1}{49^{1/2}} \] The square root of 49 is 7 (because \( 7^2 = 49 \)), so: \[ 49^{-\frac{1}{2}} = \frac{1}{7} \]
• \( 10^0 \) simplifies to 1.

Now, multiply all the simplified parts together:

\[ \frac{1}{5} \times \frac{1}{7} \times 1 = \frac{1}{5 \times 7} = \frac{1}{35} \]

The correct answer is \( \frac{1}{35} \) because: Negative exponents mean you take the reciprocal, fractional exponents correspond to roots, and anything raised to the 0 power (except 0) is 1. Multiply the simplified results to get the final fraction.