2

Is there an easy way to factor $(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)$ to get $(b+c-a) (a+b-c) (a-b+c) (a+b+c)$?

I have tried various methods such as completing the square, but I wasn't able to figure it out as it became very messy.

user26857
  • 52,094
Puzzled417
  • 6,956

2 Answers2

4

Expand and notice that it's "trying to be" $-(\pm a^2 \pm b^2 \pm c^2)^2$ for some choice of $\pm$ signs. None of the sign choices quite works, but they're each off by a square, e.g. $$ (a^2+b^2+c^2)^2-2(a^4+b^4+c^4) = 4a^2b^2 - (a^2+b^2-c^2)^2. $$ So you can factor it as the difference of two squares, and then each factor is itself a difference of two squares and thus splits further into linear forms, giving the desired factorization.

Note that the factored form is within a constant factor $16$ of Heron's formula for the square of a triangle of side lengths $a,b,c$.

Noam D. Elkies
  • 25,917
  • 1
  • 64
  • 82
  • Is there a way to see the relation to triangle area without going through the factorization? The 3d case was discussed here: http://math.stackexchange.com/questions/467952/equilateral-triangle-3a4-b4-c4-d4-a2-b2-c2-d22 – zyx Jan 21 '16 at 04:34
  • The 2-dimensional case of the Cayley-Menger determinant gives a formula for the squared area of a triangle as a polynomial in the squares of the sides. – Noam D. Elkies Jan 21 '16 at 14:32
  • I see that the Cayley-Menger determinant was already noted in a reply to that stackexchange question. In two dimensions, you might prefer starting from the formula $\frac12 ab \sin C$ for the area of a triangle, and then squaring to obtain $$ \frac14 a^2 b^2 \sin^2 C = \frac14 a^2 b^2 (1 - \cos^2 C) = \frac14 a^2 b^2 \left(1 - \frac{a^2+b^2-c^2}{2ab}\right)^{!2}. $$ The factors $(ab)^2$ cancel, and we're left with the desired polynomial in $a^2, b^2, c^2$. – Noam D. Elkies Jan 21 '16 at 15:08
2

View the expression as a polynomial in $a$. Plug in $a=b+c$. If the polynomial vanishes $(b+c-a)$ is a factor of the polynomial. Repeat for other factors. As the degree of the polynomial is $4$, it can't have more than $4$ linear factors and you're done.

Note: This method is often used to factor multivariate polynomials and is only as good as your guess.

Jack's wasted life
  • 8,849