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# Real Solution of x^3+8 == 0?

I do not understand the following:

sage: assume(x,'real')
sage: solve(x^3+8==0,x)
[]


Why does this equation have no solution? But -2 is a solution!

Thanks for help!

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## 2 Answers

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I think this is because (-1)^(1/3) is not considered to be real.

sage: solve(x^3+1==0,x)
[x == 1/2*I*(-1)^(1/3)*sqrt(3) - 1/2*(-1)^(1/3), x == -1/2*I*(-1)^(1/3)*sqrt(3) - 1/2*(-1)^(1/3), x == (-1)^(1/3)]
sage: assume(x,'real')
sage: solve(x^3+1==0,x)
[]


Note that Maxima (which does our solving) doesn't actually care about x being real, since it's a dummy variable.

(%i1) declare(x,real);
(%o1)                                done
(%i2) solve(x^3+1=0,x);
sqrt(3) %i - 1      sqrt(3) %i + 1
(%o2)         [x = - --------------, x = --------------, x = - 1]
2                   2


But when it's returned to Sage, somehow it doesn't keeps the x=-1 syntax and gets the cube root again, and it falls prey to

sage: (-1)^(1/3).n()
0.500000000000000 + 0.866025403784439*I

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## Comments

( 2011-10-18 16:47:48 +0100 )edit

An alternative approach is not to use the symbolic ring but a polynomial ring:

sage: x = polygen(RR)
sage: (x^3+8).roots()
[(-2.00000000000000, 1)]

This returns a list of roots in RR with multiplcities.

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## Stats

Asked: 2011-10-18 15:07:48 +0100

Seen: 2,600 times

Last updated: Dec 11 '13