Excuse my lack of understanding and expertise in jiyuushikan.org,but to me it would came normally that the cubic root of \$-8\$ would be \$-2\$ because \$(-2)^3 = -8\$.

You are watching: Explain why the cube root of a negative number is a negative number.

But once I checked Wolfram Alpha for \$sqrt<3>-8\$, genuine it tells me it doesn"t exist.

I came to trust Wolfram Alpha so I assumed I"d ask you guys, to define the sense of that to me.

-8 has 3 cube roots: \$ -2 \$, \$1 + i sqrt 3 \$ and \$1 - i sqrt 3 \$. So you can"t answer the question "Is \$ sqrt<3>-8 \$ real" without specifying which of them you"re talking about.

For some reason, WolframAlpha is only offering \$1 + i sqrt 3 \$ as an answer -- that looks like a bug in WolframAlpha to me.

\$egingroup\$ Actually, Wolfram is offering the correct primary cube root. jiyuushikan.orgematica is much even more oriented towards continuous jiyuushikan.org than discrete jiyuushikan.orgematics, which renders the extension of exponentiation to odd roots of negative numbers incredibly out of location. \$endgroup\$
Although it"s been two years since this question was asked, some folks can be interested to understand that this habits has actually been modified in WolframAlpha. If you ask for the cube root of an adverse number, it returns the genuine valued, negative cube root. Here, I simply asked for "cbrt -8", for example:

Keep in mind "the principal root" button. That allows you to toggle back to the original habits. Near the bottom, we still view indevelopment on all the facility roots.

We have the right to plot attributes involving the cube root and solve equations including the cube root and also it repeatedly acts actual valued. If you just kind in an equation, it will certainly resolve it, plot both sides and highlight the intersections. Here"s "cbrt(x)=sin(2x)"

Of course, you"re absolutely appropriate about \$-2\$ being a cubic root of \$-8\$.

The point could be that there are actually, three various cubic roots of \$-8\$, namely the roots of the polynomial \$x^3+8\$. One of this roots is real (\$-2\$), the various other two are complex and conjugate of each other.

I gather that Wolfram is instucted to choose one of the roots by some criteria that in this case leads to the exemption of the actual root. Maybe browsing the Wolfram site might assist knowledge what these criteria are. (My guess is that it outputs the root \$alpha=re^i heta\$ through smaller sized \$ heta\$ in the variety \$<0,2pi)\$.)

In the years considering that the question was asked and answered, Wolfram introduced the \$operatornameSurd(x,n)\$ feature (jiyuushikan.orgematica 9 circa \$2012\$, then Alpha) to designate the genuine single-valued \$n^th\$ root of \$x\$.

For instance \$sqrt<3>-8\$ and also \$sqrt<5>-243\$ outcome directly in \$-2\$ and also \$-3\$ respectively:

The \$operatornameCbrt\$ attribute disputed in Mark McClure"s answer - which had changed behavior around the same time to return the real cube root by default - shows up to be the same to \$operatornameSurd(,cdot,,3)\$:

When non-computers calculate the cube root of (-8), we can think of it as \$(-1*8)^1/3\$Then we have actually \$-1*8^1/3 = -1*2 = -2\$

Wolfram is using the polar complicated create of -8 = 8cis(π)Then the cube root of this is 2cis(π/3), which is 1 + i√3 (an different form on Wolfram)

Incidentally, if you take \$(1 + isqrt3)^3\$, you will certainly obtain -8!

Although the equation \$x^3+8 = 0\$ has actually 3 roots (actual \$-2\$ and also 2 conjugated facility roots) still the third root of \$-8\$ does not exist. Nth roots are characterized only for nonnegative genuine values. Please, think about the \$\$(-8)^1/3 = (-8)^2/6\$\$ that offers either \$64^1/6 = 2\$ or \$(sqrt -8)^2 \$ = nonsense. Let \$(-8)^1/3 = (-8)^2/6\$ you never acquire \$(-2)\$ as the outcome. Thus Nth root exists just of nonnegative genuine numbers.

Thanks for contributing an answer to jiyuushikan.orgematics Stack Exchange!

But avoid

Asking for help, clarification, or responding to various other answers.Making statements based on opinion; ago them up through referrals or personal experience.

Use jiyuushikan.orgJax to format equations. jiyuushikan.orgJax referral.

See more: 12 Reasons Why Do I Love Christmas So Much ? The Top 7 Reasons Why Christmas Is Magical