Why does 2/3 act as a multiplicative constant instead of an exponential? Moreover ##\sin \pi =0##.nomadreid said:((2/3)(cos (π) + i*sin(π))
No, it doesn't. sin(π)= 0. The other two cube roots are non-real, of course.nomadreid said:((-8)^2)^(1/3)=64^(1/3)=4
(-8)^(1/3))^2 = (-2)^2=4
but
(-8)^(2/3) = 4(-1)^(2/3) = 4((2/3)(cos (π) + i*sin(π)) and so forth, giving a complex number.
The first two do also, you just have restricted yourself to real numbers. 64 has three complex cube roots, 4 is one, the other two are non-real. 8 has 3 cube roots, one is 2 and the other two are non-real.I understand why the latter one works. Why don't the first two?
Thanks.
The expression means that we have three different expressions inside parentheses that are raised to different powers and then all of that is squared.
The value of the expression is undefined because it leads to a contradiction. The first part, "((-8)^2)^(1/3)", simplifies to 64^(1/3) which is 4. The second part, "(-8)^(2/3)", simplifies to (-64)^(1/3) which is -4. And the third part, "(-8)^(1/3)", simplifies to (-8)^(1/3) which is -2. Therefore, we have 4 <> -4 <> -2 which is a contradiction.
The expression is undefined because it leads to a contradiction. The first part, "((-8)^2)^(1/3)", simplifies to 4. The second part, "(-8)^(2/3)", simplifies to -4. And the third part, "(-8)^(1/3)", simplifies to -2. Therefore, we have 4 <> -4 <> -2 which is a contradiction.
No, the expression cannot be simplified further because it leads to a contradiction. Each part of the expression has a different value, making it impossible to simplify the expression.
Understanding this expression is important because it helps us recognize and avoid contradictions in mathematical operations. It also highlights the importance of following the correct order of operations when simplifying expressions.