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

 3 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! asked Oct 18 '11 amalea 71 ● 1 ● 7 ● 10 Jason Grout 3305 ● 7 ● 28 ● 74

1 Answer:

 3 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  posted Oct 18 '11 kcrisman 7427 ● 17 ● 76 ● 166 This is now http://trac.sagemath.org/sage_trac/ticket/11941.kcrisman (Oct 18 '11)

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Asked: Oct 18 '11

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