is there a *good* way to factorize this

[procyon]

\(\displaystyle 16u^8-64u^7 v+32u^6 v^2+160u^5 v^3+8u^4 v^4-80u^3 v^5-8u^2 v^6+8uv^7+v^8\)

I have tried starting with removing \(\displaystyle (4u^4-v^4)^2\) but the result is still messy.

The same with \(\displaystyle (2u^2-v^2)^4\)

any ideas or inspiration would be greatly appreciated

 

[galactus]

No, there is not. It does not factor nicely. Looking at it, it looks like a binomial expansion of some sort.

If they were all + signs, then it would factor to \(\displaystyle (4u^{4}+16u^{3}v+8uv^{3}+v^{4})(4u^{4}+8u^{2}v^{2}+v^{4})\)

 

[procyon]

If they were all + signs, then it would factor to \(\displaystyle (4u^{4}+16u^{3}v+8uv^{3}+v^{4})(4u^{4}+8u^{2}v^{2}+v^{4})\)

Thanks galactus, that's nice factoring on the +ve's. Just a pity it wan't that handy

I had managed this..

\(\displaystyle 3(4u^4+v^4 )^2-\left[(2u^2-v^2 )^2+4uv(2u^2-v^2 )-4u^2 v^2\right]^2-(4u^4-4u^2 v^2-v^4 )^2\)

but it seems to get like a set of matryoshka dolls, lol