Question

simplify [(1/3)^-³–(1/2)^-3]÷(1/4)^-³​

Answer 1

Answer:

$\sf \dfrac{19}{64}$

Step-by-step explanation:

We need to simplify the expression:-

$\sf \dfrac{\left[\left(\dfrac{1}{3}\right)^{-3}-\left(\dfrac{1}{2}\right)^{-3}\right]}{\left(\dfrac{1}{4}\right)^{-3}}$

This expression can be made simpler using the following relation: $\sf \dfrac{1}{x^{-a}}=x^a$

Therefore, the expression turns into:-

$\sf = \dfrac{\left[\left(3\right)^{3}-\left(2\right)^{3}\right]}{\left(4\right)^{3}} \\ \\ \\\sf =\dfrac{27-8}{64} \\ \\ \\\sf = \dfrac{19}{64}$

Note: 1 raised to any integer will always have the same value as 1. Therefore, it doesn't have to be simplified.

Explanation for using $\sf \dfrac{1}{x^{-a}}=x^a$:-

$\sf \dfrac{1}{x^a} \rightarrow \dfrac{1 \times x^0}{x^a}$

When two terms have the same base in division, their exponents are subtracted.

$\sf \dfrac{1 \times x^0}{x^a} \\ \\\sf x^{0-a} = x^{-a} \\ \\\sf \dfrac{1}{x^a}=x^{-a}$

Similarly, we can say that $\sf \dfrac{1}{x^{-a}}=x^a$

Answer 2

To simplify the expression:

$(\frac{1}{3})^{-3} - (\frac{1}{2})^{-3}] \div (\frac{1}{4})^{-3}$

We can first simplify the exponents by applying the negative exponent rule, which states that $a^{-n} = \frac{1}{a^n}$:

$[(\frac{1}{3})^{-3} - (\frac{1}{2})^{-3}] \div (\frac{1}{4})^{-3} = [(3)^{3} - (2)^{3}] \div (4)^{3}$

Next, we can simplify the expression inside the brackets:

$[(3)^{3} - (2)^{3}] \div (4)^{3} = [27 - 8] \div 64 = 19 \div 64$

Finally, we can simplify the division of the two powers of 4 by subtracting the exponent of the denominator from the exponent of the numerator:

$19 \div 64 = \frac{19}{64} \times 4^{3} = \frac{19}{64} \times 64 = 19$

Therefore, the simplified expression is 19.