Prove $(s^3+t^3+u^3+v^3)^2=9(st-uv)(tu-sv)(us-tv)$ with $s+t+u+v=0$

[anemone]

Let $s,\,t,\,u,\,v$ be real numbers such that $s+t+u+v=0$.

Prove that $(s^3+t^3+u^3+v^3)^2=9(st-uv)(tu-sv)(us-tv)$.

 

[kaliprasad]

Let $s,\,t,\,u,\,v$ be real numbers such that $s+t+u+v=0$.

Prove that $(s^3+t^3+u^3+v^3)^2=9(st-uv)(tu-sv)(us-tv)$.

Spoiler

from the above
$(s+t) = -(u+v)\cdots(1)$
also
$s+t+ u = -v \cdots(2)$
cube both sides of (1) to get
$(s+t)^3 = - (u+v)^3$
or $s^3+t^3 + 3st(s+t) = - (u^3+v^3 + 3uv(u+v)$
or $s^3+t^3+ u^3+v^3 = -(3st(s+t) + 3uv(u+v))= -3(st(s+t) - uv(s+t))$ using (1)
or $s^3+t^3+u^3+v^3 = -3(st-uv)(s+t)\cdots(3)$
by symmetry we can show that
$s^3+t^3+u^3+v^3 = -3(su-tv)(s+u)\cdots(4)$
by multiplying (3) with (4) we get
$(s^3+t^3+u^3+v^3)^2 = 9(st-uv)(su-tv)((s+t)(s+u))$
= $9(st-uv)(su-tv)(s^2+ st + su + ut)$
= $9(st-uv)(su-tv)(s(s+t+u) + ut)$
= $9(st-uv)(su-tv)(-vs + ut)$ using (2)
= $9(st-uv)(ut-sv)(su-tv)$

Proved

 

[Greg]

Let $s,\,t,\,u,\,v$ be real numbers such that $s+t+u+v=0$.

Prove that $(s^3+t^3+u^3+v^3)^2=9(st-uv)(tu-sv)(us-tv)$.

Spoiler

$$\text{L.H.S:}$$

$$(s^3+t^3+u^3+v^3)^2$$

$$=[(s+t)(s^2-st+t^2)+(u+v)(u^2-uv+v^2)]^2$$

$$=[(s+t)(s^2-st+t^2)-(s+t)(u^2-uv+v^2)]^2$$

$$=(s+t)^2(s^2-u^2+t^2-v^2-st+uv)^2$$

$$=(s+t)^2[(s-u)(s+u)+(t-v)(t+v)-st+uv]^2$$

$$=(s+t)^2[-(s-u)(t+v)-(t-v)(s+u)-st+uv]^2$$

$$=(s+t)^2(-st-sv+tu+uv-st-tu+sv+uv-st+uv)^2$$

$$=(s+t)^2(-3st+3uv)^2$$

$$=9(st-uv)^2(s+t)^2$$

$$(st-uv)(s+t)^2=st(u+v)^2-uv(s+t)^2=stu^2+2stuv+stv^2-s^2uv-2stuv-t^2uv=stu^2-t^2uv-s^2uv+stv^2$$

$$\text{R.H.S:}$$

$$9(st-uv)(tu-sv)(su-tv)$$

$$(tu-sv)(su-tv)=stu^2-t^2uv-s^2uv+stv^2$$

(as required).

 

[anemone]

Thanks to kaliprasad and greg1313 for participating in this challenge! :)

 

[kaliprasad]

Spoiler

$$(st-uv)(s+t)^2=st(u+v)^2-uv(s+t)^2=stu^2+2stuv+stv^2-s^2uv-2stuv-t^2uv=stu^2-t^2uv-s^2uv+stv^2$$

Spoiler

smart reduction I struggled here before I changed my method