What's the difference between $-81^{3/2}$ & $(-81)^{3/2}$?

Question

Calculating $81^{3/2}$, I got $729$ (not saying it is correct, but I am trying :) ). Would $-81^{3/2}$ just be the opposite ($-729$) and does it make a difference if $-81$ was placed inside a pair of parentheses $(-81)^{3/2}$?

Answer 1

We can think of the $-$ sign as the following words: "the opposite of".

In this case, $-1$ is the opposite of $1$, which just means it's the number that, when added to $1$, results in $0$.

Similarly, $-\pi$ is the opposite of $\pi$, the number that, when added to $\pi$, results in $0$.

In your question, $-81^{3/2}$ is a negative number, the number that, when added to $81^{3/2}$, results in $0$. You can simplify $81^{3/2}=729$ to see that $-81^{3/2}$ then must be $-729$.

However, $(-81)^{3/2}$ has the $-$ sign inside the parentheses. This means that the "opposite of" quality applies to $81$ and THEN we apply an exponent. Using properties of exponents, we do get that this is equal to $(-1)^{3/2}(81)^{3/2}$. If you are unfamiliar with complex numbers, then just saying that you can't take the square root of the $-1$ is sufficient for this, and is enough to see that this is very different than the original $-81^{3/2}$.

Answer 2

I'm also new to Mathematics but I will try and explain in simple terms:

Using exponent rules we can say that $\left(-81\right)^{\frac{3}{2}}$ is equal to $\sqrt[2]{\left(-81\right)^{3}}$. In this case $\left(-81\right)^{3}$ will give us a negative number, so calculating $\left(-81\right)^{\frac{3}{2}}$ is only possible with the use of imaginary numbers.

It does make a difference if you put it in parentheses. If you type $-81^{\frac{3}{2}}$ in a calculator it will give you the anwer $-729$; this is only because it determines the expression as $-\left(81^{\frac{3}{2}}\right)$ meaning it will first evaluate the expression $\sqrt[2]{81^{3}}$ which will result in a positive number and then multiply it by $-1$.

More info: https://www.khanacademy.org/math/cc-eighth-grade-math/cc-8th-numbers-operations/exponents-with-negative-bases/v/exponents-with-negative-bases

Cheers, Tom

(I am a math 'noob' so if I've missed something please let me know.)