FMU Question 1 Use exponent rules to simplify the following, giving all answers with positive exponents. (a) \( 6^{3} \times 6^{-5} \times 6^{-2} \) \( 3+(5)+(-2)=3-5-2=3-7=\frac{1}{64}=6^{-4} \times 6^{-2}=\frac{1}{6} \) \( 6^{3} \times 6^{-5} \times 6^{-2}= \) (b) \( \quad-5\left(a^{3}\right)^{0}-3+4^{0} \)

Let's simplify each part step by step. ### Part (a) We start with the expression: \[ 6^{3} \times 6^{-5} \times 6^{-2} \] Using the exponent rule \( a^m \times a^n = a^{m+n} \), we can combine the exponents: \[ 6^{3 + (-5) + (-2)} = 6^{3 - 5 - 2} = 6^{-4} \] To express this with a positive exponent, we use the rule \( a^{-m} = \frac{1}{a^m} \): \[ 6^{-4} = \frac{1}{6^{4}} \] Now, we calculate \( 6^4 \): \[ 6^4 = 6 \times 6 \times 6 \times 6 = 1296 \] Thus, we have: \[ 6^{-4} = \frac{1}{1296} \] So the final answer for part (a) is: \[ 6^{3} \times 6^{-5} \times 6^{-2} = \frac{1}{1296} \] ### Part (b) We simplify the expression: \[ -5\left(a^{3}\right)^{0}-3+4^{0} \] First, we apply the exponent rule \( a^0 = 1 \) for any non-zero \( a \): \[ \left(a^{3}\right)^{0} = 1 \quad \text{and} \quad 4^{0} = 1 \] Substituting these values into the expression gives: \[ -5(1) - 3 + 1 \] Now, we simplify: \[ -5 - 3 + 1 = -8 + 1 = -7 \] So the final answer for part (b) is: \[ -5\left(a^{3}\right)^{0}-3+4^{0} = -7 \] ### Summary of Answers (a) \( 6^{3} \times 6^{-5} \times 6^{-2} = \frac{1}{1296} \) (b) \( -5\left(a^{3}\right)^{0}-3+4^{0} = -7 \)