Question

$\frac{4^6}{4^2 \times 4} \times \frac{4^3}{4^2 \times 4}$

Explanation:

Step1: Simplify denominators using exponent rules

Recall that \(a^m \times a^n = a^{m + n}\). For the denominator \(4^2 \times 4\), we have \(4^2\times4^1=4^{2 + 1}=4^3\). So both fractions have denominator \(4^3\). The expression becomes \(\frac{4^6}{4^3}\times\frac{4^3}{4^3}\).

Step2: Simplify each fraction

For \(\frac{4^6}{4^3}\), use the rule \(\frac{a^m}{a^n}=a^{m - n}\), so \(\frac{4^6}{4^3}=4^{6-3}=4^3\). For \(\frac{4^3}{4^3}\), this simplifies to \(1\) (since any non - zero number divided by itself is \(1\)).

Step3: Multiply the results

Now we multiply \(4^3\times1\). And \(4^3 = 4\times4\times4=64\).

Answer:

\(64\)