Question

question 2
without using a calculator, find the value of ((3 \times 2^2)^{-3} - 3(2^3)^{-2} + \frac{3^{-3}(2^3)^{-2}}{(5^2 + 1)^{-1}}
give your answer as a fraction without spaces.

Explanation:

Step1: Simplify each term using exponent rules

First, recall the exponent rules: \( (ab)^n = a^n b^n \), \( (a^m)^n = a^{mn} \), \( a^{-n} = \frac{1}{a^n} \), and \( (a + b)^{-1}=\frac{1}{a + b} \).

Term 1: \( (3\times2^2)^{-3} \)

First, calculate \( 2^2 = 4 \), then \( 3\times4 = 12 \). So \( (12)^{-3}=\frac{1}{12^3}=\frac{1}{1728} \).

Term 2: \( -3(2^3)^{-2} \)

Calculate \( 2^3 = 8 \), then \( (8)^{-2}=\frac{1}{8^2}=\frac{1}{64} \). Multiply by -3: \( -3\times\frac{1}{64}=-\frac{3}{64} \).

Term 3: \( \frac{3^{-3}(2^3)^{-2}}{(5^2 + 1)^{-1}} \)

First, simplify numerator and denominator separately.
- Numerator: \( 3^{-3}=\frac{1}{3^3}=\frac{1}{27} \), \( (2^3)^{-2}=2^{-6}=\frac{1}{2^6}=\frac{1}{64} \). Multiply them: \( \frac{1}{27}\times\frac{1}{64}=\frac{1}{1728} \).
- Denominator: \( 5^2 + 1 = 25 + 1 = 26 \), so \( (26)^{-1}=\frac{1}{26} \).
Now, divide the numerator by the denominator: \( \frac{\frac{1}{1728}}{\frac{1}{26}}=\frac{1}{1728}\times26=\frac{26}{1728}=\frac{13}{864} \).

Step2: Combine all terms

Now, add the three terms together:
\( \frac{1}{1728}-\frac{3}{64}+\frac{13}{864} \)
First, find a common denominator. The least common multiple of 1728, 64, and 864. Notice that 1728 is a multiple of 64 (1728 ÷ 64 = 27) and 864 (1728 ÷ 864 = 2). So common denominator is 1728.
- \( -\frac{3}{64}=-\frac{3\times27}{64\times27}=-\frac{81}{1728} \)
- \( \frac{13}{864}=\frac{13\times2}{864\times2}=\frac{26}{1728} \)
Now, substitute back:
\( \frac{1}{1728}-\frac{81}{1728}+\frac{26}{1728}=\frac{1 - 81 + 26}{1728}=\frac{-54}{1728} \)
Simplify the fraction: divide numerator and denominator by 18: \( \frac{-54\div18}{1728\div18}=\frac{-3}{96} \), divide by 3 again: \( \frac{-1}{32} \)

Wait, let's check the calculations again for errors.

Wait, let's re - calculate Term 3:

Numerator: \( 3^{-3}(2^3)^{-2}=3^{-3}\times2^{-6}=\frac{1}{3^3}\times\frac{1}{2^6}=\frac{1}{27}\times\frac{1}{64}=\frac{1}{1728} \)

Denominator: \( (5^2 + 1)^{-1}=(25 + 1)^{-1}=26^{-1}=\frac{1}{26} \)

So \( \frac{\frac{1}{1728}}{\frac{1}{26}}=\frac{26}{1728}=\frac{13}{864} \) (correct)

Term 1: \( (3\times2^2)^{-3}=(3\times4)^{-3}=12^{-3}=\frac{1}{12^3}=\frac{1}{1728} \) (correct)

Term 2: \( -3(2^3)^{-2}=-3\times(8)^{-2}=-3\times\frac{1}{64}=-\frac{3}{64} \). Convert to denominator 1728: \( -\frac{3}{64}=-\frac{3\times27}{64\times27}=-\frac{81}{1728} \) (correct)

Now, \( \frac{1}{1728}-\frac{81}{1728}+\frac{13}{864} \)

Convert \( \frac{13}{864} \) to \( \frac{26}{1728} \) (since 864×2 = 1728, 13×2 = 26)

So \( \frac{1-81 + 26}{1728}=\frac{-54}{1728} \)

Simplify \( \frac{-54}{1728} \): divide numerator and denominator by 18: \( \frac{-3}{96} \), divide by 3: \( \frac{-1}{32} \)

Wait, but let's check the original expression again. Maybe I misread the expression. Let's re - examine the original problem:

The expression is \( (3\times2^2)^{-3}-3(2^3)^{-2}+\frac{3^{-3}(2^3)^{-2}}{(5^2 + 1)^{-1}} \)

Wait, in the third term, is it \( (2^3)^{-2} \) or \( (2^2)^{-3} \)? Wait, the original problem: \( \frac{3^{-3}(2^3)^{-2}}{(5^2 + 1)^{-1}} \)

Wait, let's recalculate Term 3:

\( 3^{-3}=\frac{1}{27} \), \( (2^3)^{-2}=2^{-6}=\frac{1}{64} \), so numerator is \( \frac{1}{27}\times\frac{1}{64}=\frac{1}{1728} \)

Denominator: \( (5^2 + 1)^{-1}=\frac{1}{26} \)

So \( \frac{\frac{1}{1728}}{\frac{1}{26}}=\frac{26}{1728}=\frac{13}{864} \)

Term 1: \( (3\times2^2)^{-3}=(12)^{-3}=\frac{1}{1728} \)

Term 2: \( -3\times(2^3)^{-2}=-3\times\frac{1}{64}=-\frac{3}{64} \)

Now, let's convert all terms to 1728 denominator:

\( \frac{1}{1728}-\frac{3\times27}{64\times27}+\frac{13\times2}{864\times2}=\frac{1}{1728}-\frac{81}{1728}+\frac{26}{1728}=\frac{1 - 81+26}{1728}=\frac{-54}{1728}=\frac{-1}{32} \)

Wait, but let's check the arithmetic again: 1 - 81 is - 80, - 80+26 is - 54. Yes. And - 54/1728: divide numerator and denominator by 18: - 3/96, divide by 3: - 1/32.

Answer:

\(-\frac{1}{32}\)