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

In summary, the conversation discussed an equation that needs to be proven, which is (s^3+t^3+u^3+v^3)^2=9(st-uv)(tu-sv)(us-tv). The variables in the equation are s, t, u, and v. The condition for the equation to hold true is that s+t+u+v=0. This equation is significant because it is a special case of a well-known identity and has various applications in algebra and number theory. The process for proving the equation involves simplifying both sides and showing their equality using mathematical properties and rules.
  • #1
anemone
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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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  • #2
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
 
  • #3
anemone said:
Let $s,\,t,\,u,\,v$ be real numbers such that $s+t+u+v=0$.

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

\(\displaystyle \text{L.H.S:}\)

\(\displaystyle (s^3+t^3+u^3+v^3)^2\)

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

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

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

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

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

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

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

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

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

\(\displaystyle \text{R.H.S:}\)

\(\displaystyle 9(st-uv)(tu-sv)(su-tv)\)

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

(as required).
 
  • #4
Thanks to kaliprasad and greg1313 for participating in this challenge! :)
 
  • #5
greg1313 said:
\(\displaystyle (st-uv)(s+t)^2=st(u+v)^2-uv(s+t)^2=stu^2+2stuv+stv^2-s^2uv-2stuv-t^2uv=stu^2-t^2uv-s^2uv+stv^2\)
smart reduction I struggled here before I changed my method
 

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

What is the equation to be proven?

The equation to be proven is:
(s^3+t^3+u^3+v^3)^2=9(st-uv)(tu-sv)(us-tv)

What are the variables in the equation?

The variables in the equation are s, t, u, and v.

What is the condition for the equation to hold true?

The condition for the equation to hold true is that s+t+u+v=0.

What is the significance of the equation?

The equation is significant because it is a special case of the well-known identity (a^3+b^3)^2=9ab(a-b)^2(a^2+ab+b^2), and it has various applications in algebra and number theory.

What is the process for proving the equation?

The process for proving the equation involves simplifying both sides of the equation using algebraic manipulations and substitution of the given condition (s+t+u+v=0). Then, showing that both sides are equal using mathematical properties and rules.

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