Solving an cubic equation

Question

This comes from a chemistry question but it is the maths I am struggling with. I solved the determinant of a $3 \times 3$ matrix to get:

$$(a-E)^3 - 3B^2(a-E) + 2B^3 = 0$$

I need to solve this in terms of $a$ and $B$. So for example $E = a + 2B$.

If you break it down into brackets, (.....)(.....) then I would be able to see how to make it zero as if one bracket equals zero the whole equation will. However I am unsure of how to do this.

Answer 1

Let $x=a-E$ then

$$(a-E)^3 - 3B^2(a-E) + 2B^3 = 0 \iff x^3-3B^2x+2B^3=0$$

and

$$x^3-3B^2x+2B^3=x^3-B^2x-2B^2x+2B^3=x(x-B)(x+B)-2B(x-B)=(x-B)(x^2+xB-2B)=(x-B)(x-B)(x+2B)=(x-B)^2(x+2B)=0$$

thus

• $x-B=0\implies E=a-B$
• $x+2B=0\implies E=a+2B$

Answer 2

your equation can be factorized into $$(a+2 B-x) (-a+B+x)^2=0$$ i have changed the $E$ to $x$