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

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The equation $(s^3+t^3+u^3+v^3)^2=9(st-uv)(tu-sv)(us-tv)$ is to be proven under the condition that $s+t+u+v=0$. Participants discuss the implications of this condition on the cubic terms and the products of the variables. The proof involves algebraic manipulation and the application of symmetric sums. The challenge highlights the relationship between the sums of cubes and the products of pairs of variables. The discussion acknowledges contributions from users kaliprasad and greg1313.
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Let $s,\,t,\,u,\,v$ be real numbers such that $s+t+u+v=0$.

Prove that $(s^3+t^3+u^3+v^3)^2=9(st-uv)(tu-sv)(us-tv)$.
 
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anemone said:
Let $s,\,t,\,u,\,v$ be real numbers such that $s+t+u+v=0$.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

(as required).
 
Thanks to kaliprasad and greg1313 for participating in this challenge! :)
 
greg1313 said:
$$(st-uv)(s+t)^2=st(u+v)^2-uv(s+t)^2=stu^2+2stuv+stv^2-s^2uv-2stuv-t^2uv=stu^2-t^2uv-s^2uv+stv^2$$
smart reduction I struggled here before I changed my method
 
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