Solve: (-8)^(4/3)?

That would be cube root of (-8)^(4) or (-2)^4 = 16.

Notice that if you'd taken the cube root of 8^4 you'd arrive at the same conclusion. that's because multiplication is associative.

You can take cube roots (or any odd roots) of negative numbers and remain in the real number system. It is only square roots and other even roots of negative numbers which require complex numbers.

In addition, fractional exponents are defined as roots. To understand why this is, consider the laws of exponents. Since (a^m)^n = a^(mn), and a^1 = 1, it follows that a^(1/m) = the mth root of a. So, for example, 2^(1/2) = square root of 3, 6^(1/3) = cube root of 6, 12^(1/5) = fifth root of 12, etc.

This is the same thing as the cube root of -8, which is -2, raised to the fourth power.

(-8)^(4/3) = [(-8)^1/3]^4 = (-2)^4 = 16

16

(-8)^(4/3)=(-2)^4=16

= ((-8)^(1/3))^(4)

= ((-2*-2*-2)^(1/3))^(4)

= (-2)^(4)

= -2*-2*-2*-2

= 4*4

= 16

-8^4/3=

=the cuberoot of -8^4

=the cuberoot of -4096

= -16

(-8)^(4/3)

=(-2)^4

=16

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In real domain, this is the only answer. However, in complex domain, you can have three answers in its major branch.