Question

What are the solutions of the equation x6 + 6x3 + 5 = 0? Use factoring to solve.

Answer 1

Consider the equation $x^6+6x^3+5=0$.

First, you can use the substitution $t=x^3$, then $x^6=(x^3)^2=t^2$ and equation becomes $t^2+6t+5=0$. This equation is quadratic, so

$D=6^2-4\cdot 5\cdot 1=36-20=16=4^2,\ \sqrt{D}=4,\\ \\ t_{1,2}=\dfrac{-6\pm 4}{2} =-5,-1$.

Then you can factor this equation:

$(t+5)(t+1)=0$.

Use the made substitution again:

$(x^3+5)(x^3+1)=0$.

You have in each brackets the expression like $a^3+b^3$ that is equal to $(a+b)(a^2-ab+b^2)$. Thus,

$x^3+5=(x+\sqrt[3]{5})(x^2-\sqrt[3]{5}x+\sqrt[3]{25}) ,\\x^3+1=(x+1)(x^2-x+1)$

and the equation is

$(x+\sqrt[3]{5})(x^2-\sqrt[3]{5}x+\sqrt[3]{25})(x+1)(x^2-x+1)=0$.

Here $x_1=-\sqrt[3]{5} , x_2=-1$ and you can sheck whether quadratic trinomials have real roots:

1. $D_1=(-\sqrt[3]{5}) ^2-4\cdot \sqrt[3]{25}=\sqrt[3]{25} -4\sqrt[3]{25} =-3\sqrt[3]{25}$.

2. $D_2=(-1)^2-4\cdot 1=1-4=-3$.

This means that quadratic trinomials don't have real roots.

Answer: $x_1=-\sqrt[3]{5} , x_2=-1$

If you need complex roots, then

$x_{3,4}=\dfrac{\sqrt[3]{5}\pm i\sqrt{3\sqrt[3]{25}}}{2} ,\\ \\x_{5,6}=\dfrac{1\pm i\sqrt{3}}{2}$.