The determinant is homogeneous in a, b and c, so let b=xa, c=ya, where x and y are constants.
The determinant can then be written
| a²(x+y)² a² a² |
| a²x² a²(1+y)² a²x² |
| a²y² a²y² a²(1+x)² |
a² can be taken outside the determinant as a⁶D, where D=
| (x+y)² 1 1 |
| x² (1+y)² x² | =
| y² y² (1+x)² |
(x+y)²[(1+x)²(1+y)²-x²y²]
-[x²(1+x)²-x²y²]
+[x²y²-y²(1+y)²]=
(x+y)²(1+x+y)(1+x+y+2xy)
-x²(1+x+y)(1+x-y)
-y²(1+x+y)(1+y-x)=
(1+x+y)[(x+y)²(1+x+y+2xy)
-x²(1+x-y)-y²(1+y-x)]=
(1+x+y)[(x²+2xy+y²)(1+x+y+2xy)
-x²-x³+x²y-y²-y³+xy²]=
(1+x+y)(x²+x³+x²y+2x³y+2xy+2x²y+2xy²+4x²y²+y²+xy²+y³+2xy³
-x²-x³+x²y-y²-y³+xy²)=
(1+x+y)(4x²y+2x³y+2xy+4xy²+4x²y²+2xy³)=
2xy(1+x+y)(2x+x²+1+2y+2xy+y²)=
2xy(1+x+y)(x²+2x+1+y²+2y(x+1))=
2xy(1+x+y)((x+1)²+2y(x+1)+y²)=
2xy(1+x+y)(x+1+y)²=
2xy(1+x+y)³.
Therefore, a⁶D=2a⁶xy(1+x+y)³.
Since x=b/a and y=c/a:
a⁶D=2a⁴bc(1+b/a+c/a)³=2a⁴bc[(a+b+c)/a]³=
2abc(a+b+c)³.