2016/111) if $s+t+u+v = 0$

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

Solution

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