Answer:
[tex]B)\quad x=-\sqrt[3]{5}\quad\text{and}\quad x=-1[/tex]
Step-by-step explanation:
The equation is a quadratic in x³, so can be factored the same way the quadratic x² +6x +5 = 0 would be factored.
... (x³ +1)(x³ +5) = 0
The solutions are the values of x that make the factors be zero. That is, the expression resolves to two cubic equations. (Each of those has 1 real root and 2 complex roots. We'll ignore the complex roots.)
For the first factor:
... x³ +1 = 0
... x³ = -1
... x = ∛(-1) = -1 . . . . . real root (complex roots ignored)
For the second factor:
... x³ +5 = 0
... x³ = -5
... x = ∛(-5) = -∛5 . . . . . real root (complex roots ignored)
The real solutions to the equation are x = -∛5 and x = -1.
