11 (a) Find the value of $\sqrt{81} \times 10^8$ (b) Find the value of $64^{-\frac{1}{3}}$ (c) Write $\frac{3^{n}}{9^{n-1}}$ as a power of 3. - Edexcel - GCSE Maths - Question 11 - 2020 - Paper 1

First, find the cube root of 64.

$\sqrt[3]{64} = 4$

Since we have an exponent of $-\frac{1}{3}$, we take the reciprocal:

$64^{-\frac{1}{3}} = \frac{1}{4}$.

Rewrite 9 as $3^2$:

$\frac{3^{n}}{(3^2)^{n-1}}$

This simplifies to:

$\frac{3^{n}}{3^{2(n-1)}} = 3^{n - 2(n-1)} = 3^{n - 2n + 2} = 3^{2 - n}$

Thus, the final answer is $3^{2 - n}$.