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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.

Raskolnikov
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Henry Cooper
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2 Answers2

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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

  • $x-B=0\implies E=a-B$
  • $x+2B=0\implies E=a+2B$
user
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    The transformation $x = a - E$ is a useful simplification, but since cubics (other than sum/difference of cubes) is not a standard school factorization topic, it would probably be helpful to remind the OP that the factorization can be discovered by the precalculus rational roots theorem. In this case, the trials for $p/q$ are divisors of $2B^3$ divided by divisors of $1,$ and thus are $\pm 2B^3,$ $\pm B^3,$ $\pm 2B^2,$ $\pm B^2,$ etc. (moments later) While I was writing this, you provided an alternate way of obtaining the factorization. – Dave L. Renfro Feb 08 '18 at 10:37
  • @DaveL.Renfro You are completely right, thanks for the pointing out! – user Feb 08 '18 at 10:42
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your equation can be factorized into $$(a+2 B-x) (-a+B+x)^2=0$$ i have changed the $E$ to $x$

Dr. Sonnhard Graubner
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