Question

how do I turn bc^2-ac^2-b^2c+ab^2+a^2c-a^2b into (b-a)(c-a)(c-b)

Answer 1

\[bc ^{2}-b ^{2}c-ac ^{2}+ab ^{2}+a ^{2}c-a ^{2}b\]

Answer 2

if it helps, it's the vandermonde determinant [1 a a^2],[1 b b^2], [1 c c^2], and I have to show that it equals (b-a)(c-a)(c-b) for all scalars a,b,c

Answer 3

any suggestions?

Answer 4

You need to add a 0 to make it factor nicely. In particular -bca + acb.

\[bc^2 - b^2c - bca + b^2a = b(c^2 - bc -ca + ba)\]
\[-ac^2 + acb + a^2c - a^2b = -a(c^2 - cb - ca + ba)\]
\[c^2 - cb - ca + ba = (c-a)(c-b)\]

So taking all the terms together you get (b-a)(c-a)(c-b)

Answer 5

thanks