Is there a number exactly 1 more than its cube?

Et x be "a number" 
then its cube is x^3 
one more than its cube is 1+x^3 
so the equation should be 
x = x^3 + 1 
"a number" = "one more than its cube" 
if you rearranged it to the form you will have, it would be 
x MINUS 1 = x^3, rather than plus 
anyway, x is approximate -1.325. you get this by solving x^3 - x + 1 = 0 
you know you will find a real solution because a consequence of the IVT is that every polynomial of odd degree has at least one real root. check your text book or use google to find a proof of that.