factorizing the given polynomial.

Question

factorize $(bc-a^2)(ac-b^2)-(c^2-ab)^2$. The one method I know is to something like let $ a=x$ and if equation becomes zero then $a-x$ is a root but how to exactly think?

Answer 1

I think it's better to do this using determinants,the given equation can be converted to the following matrix: $$\begin{pmatrix}{bc-a^2} & {ab-c^2}\\\ {ab-c^2} & {ac-b^2}\end{pmatrix}$$ and we just need to find the determinant of this matrix to the given equation.This can be easily done with help of row and column transformations $$\det(\begin{pmatrix}{bc-a^2} & {ab-c^2}\\\ {ab-c^2} & {ac-b^2}\end{pmatrix})=({bc-a^2})({ac-b^2})-({ab-c^2})^2$$and using $$C_1\to C_1-C_2$$we get $$\det(\begin{pmatrix}{(c-a)(a+b+c)} & {ab-c^2}\\\ {(b-c)(a+b+c)} & {ac-b^2}\end{pmatrix})$$ and taking a+b+c common from $C_1$ we get $$(a+b+c)\det(\begin{pmatrix}{c-a} & {ab-c^2}\\\ {b-c} & {ac-b^2}\end{pmatrix})$$ and now evaluating the det value we get $$({bc-a^2})({ac-b^2})-({ab-c^2})^2=(a+b+c)(c)(ab+ac+bc-a^2-b^2-c^2)$$ which simplifies to $$({bc-a^2})({ac-b^2})-({ab-c^2})^2=-(a+b+c)(c)(\frac{1}{2})((a-b)^2+(b-c)^2+(c-a)^2)$$

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And a simpler way is to just simply it as given:$$({bc-a^2})({ac-b^2})-({ab-c^2})^2=abc^2-b^3c-a^3c+a^2b^2-c^4-a^2b^2+2abc^2$$now taking -c common from the whole equation gives us $$(-c)(abc-b^3-a^3-c^3+2abc)=(-c)(-3abc+a^3+b^3+c^3)=(-c)(\frac{1}{2}((a-b)^2+(b-c)^2+(c-a)^2)(a+b+c)$$

Answer 2

Note that $c=0$ is clearly a solution. Notwithstanding, expand out into $$abc^2 - a^3c - b^3c + a^2b^2 - c^4 + 2abc^2 - a^2b^2$$ $$3abc^2 - a^3c - b^3c - c^4$$ $$c(3abc - a^3 - b^3 - c^3)$$ Note that $a+b+c = 0$ is a solution and then factor it again. $$-c(a+b+c)(a^2 + b^2 + c^2 - ab - bc - ac)$$