11 (a) Find the value of \( \sqrt{81} \times 10^8 \) (b) Find the value of \( 64^{-\frac{1}{3}} \) (c) Write \( \frac{3^{n-1}}{9^{n-1}} \) as a power of 3 (Total for Question 11 is 6 marks) - Edexcel - GCSE Maths - Question 15 - 2020 - Paper 1

First, we calculate the cube root of 64. Since ( 64 = 4^3 ), it follows that ( \sqrt[3]{64} = 4 ). Therefore,

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

The value is ( \frac{1}{4} ).

First, we note that ( 9 = 3^2 ). Hence,

$9^{n-1} = (3^2)^{n-1} = 3^{2(n-1)}.$

This implies that

$\frac{3^{n-1}}{9^{n-1}} = \frac{3^{n-1}}{3^{2(n-1)}} = 3^{(n-1) - 2(n-1)} = 3^{(n-1 - 2(n-1))} = 3^{n-1 - 2n + 2} = 3^{-n + 1}.$

Thus, ( \frac{3^{n-1}}{9^{n-1}} = 3^{-n + 1}. $$