Question

With steps please?

Answer 1

`z_1^3 z_2^6 = 64`

Explanation:

If `n` is any integer, then for all `a, b` we have:

`(a^b)^n = a^(b * n)`

Euler's identity:

`e^(ipi) + 1 = 0`

Given:

`{ (z_1 = e^(ipi/3)), (z_2 = 2e^(ipi/6)) :}`

Then:

`z_1^3 z_2^6 = (e^(ipi/3))^3 * (2 e^(ipi/6))^6`

`color(white)(z_1^3 z_2^6) = e^((ipi/3) * 3) * 2^6 * e^((ipi/6) * 6)`

`color(white)(z_1^3 z_2^6) = e^(ipi) * 64 * e^(ipi)`

`color(white)(z_1^3 z_2^6) = (-1) * 64 * (-1)`

`color(white)(z_1^3 z_2^6) = 64`

Footnote

The identity `(a^b)^c = a^(bc)` does not always hold.

For example, if `a=-1`, `b=2` and `c=1/2` then:

`((-1)^2)^(1/2) = 1^(1/2) = 1 != -1 = (-1)^1 = (-1)^(2 * 1/2)`