What are the solutions of the equation x^6 + 6x^3 + 5 = 0? Use factoring to solve A) x = ^3√5 and x = 1B) x = -^3√5 and x = -1C) x = ^3√5 and x = -1D) x = -^3√5 and x = 1

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) = 0The 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.