Question

Evaluate the following:

`(27)^(4/3) - 5^0 xx (1/9)^(-3/2) + (81)^(1/2)`

Answer 1

Given expression: `(27)^(4/3) - 5^0 xx (1/9)^(-3/2) + (81)^(1/2)`

Step-wise calculation:

1. Calculate `(27)^(4/3)`:

Note 27 = 3³.

`(27)^(4/3) = (3^3)^(4/3)`

`(27)^(4/3) = 3^(3 xx 4/3)`

`(27)^(4/3) = 3^4`

`(27)^(4/3) = 81`

2. Calculate 5⁰:

Any non-zero number to the power 0 is 1.

So, 5⁰ = 1.

3. Calculate `(1/9)^(-3/2)`:

Note 9 = 3²,

So `1/9 = 9^-1 = (3^2)^-1`

`1/9 = 9^-1 = 3^-2`

Then `(1/9)^(-3/2) = (3^-2)^(-3/2)`

`(1/9)^(-3/2) = 3^(-2 xx - 3/2)`

`(1/9)^(-3/2) = 3^3`

`(1/9)^(-3/2) = 27`

4. Thus, `5^0 xx (1/9)^(-3/2) = 1 xx 27`

`5^0 xx (1/9)^(-3/2) = 27`

5. Calculate `(81)^(1/2)`:

Note 81 = 9² = 3⁴.

`(81)^(1/2) = (3^4)^(1/2)`

`(81)^(1/2) = 3^(4 xx 1/2)`

`(81)^(1/2) = 3^2`

`(81)^(1/2) = 9`

6. Now substitute all values back into the original expression:

81 – 27 + 9

7. Simplify:

81 – 27 = 54 

54 + 9 = 63