why are these answers no real root?

(-81)^3/4

-5X^6=320

Because they are both the even-numbered root of a negative number. 81^3 is negative and to take the fourth root of that, you have to take a square root twice.

thanks

To determine whether the equations have real roots, we can analyze the expression within the parentheses in the exponent and the equation itself.

For the expression (-81)^(3/4), we can simplify it step by step:
1. Start by evaluating the root: √(-81) = √(-1 * 81) = 9i.
2. Now, let's raise 9i to the power of 3/4: (9i)^(3/4).
To simplify this, we convert the complex number from polar form (r ∠ θ) to exponential form (re^iθ):
9i = 9 * e^(iπ/2) (where iπ/2 represents a quarter-turn counterclockwise on the complex plane).
Now, raising the exponential form to the power of 3/4:
(9 * e^(iπ/2))^(3/4) = 9^(3/4) * e^[(3/4) * (iπ/2)].
3. Evaluating 9^(3/4) = (9^(1/4))^3 = (3)^3 = 27.
Therefore, the expression simplifies to 27 * e^[(3/4) * (iπ/2)].

Considering the equation -5x^6 = 320:
1. Divide both sides by -5: x^6 = -64.
2. We need to find the sixth root of -64.
For this equation, there is a real root because (-64)^(1/6) = -2, which satisfies the equation x^6 = -64.

In summary:
- The expression (-81)^(3/4) has no real root because it involves complex numbers.
- The equation -5x^6 = 320 has at least one real root, which is x = -2.