Factoring

[thorn mirage]

I need to use this so what I was trying to do was manipulate the formula so it fits the a^3 + b^3 + c^3 - 3abc

but I don't really know how to

[quiet plover]

You can factor by m+n-33

$m^3+n^3+99mn - 33^3= (m + n - 33) (m^2 - m n + 33 m + n^2 + 33 n + 1089)$

[willow furnace]

Daddy_314

[quiet plover]

The aim would be to write $x^3+y^3+ z^3$ in terms of the elementary symmetric polynomials:
$x^3+y^3+z^3 = (x + y + z)^3 - 3 (x y + x z + y z) (x + y + z) + 3 x y z$

Which means $x^3+y^3+z^3-3xyx$ can be factored by $x+y+z$

Then you can adapt this to m, n and -33

[willow furnace]

Daddy_314

[thorn mirage]

@quiet plover

I don't really understand

a^3 + b^3 + c^3 is what we are trying to get but whats the c here?

[quiet plover]

m and n satisfy
$m^3 + n^3 - 33^3 + 99mn = 0$

[willow furnace]

Daddy_314

[quiet plover]

Which means
$m^3 + n^3 + (-33)^3 - 3 \times (-33) \times mn = 0$

[willow furnace]

Daddy_314

[quiet plover]

So c = -33 basically

[thorn mirage]

oh ok

[thorn mirage]

what would be next step from here?

i get it now