Show that $q^4+2pq^2 +p^2 = 2pq -(pq)^2 -1$ becomes $p^3+q^3+3pq-1=0$.

Question

I know that these two are exactly the same equation but I can't seem to prove it.

You are also given that $p+q=1$.

This is a follow up from a similar question.

Answer 1

As lab said.

$$(q^2+p)^2+(pq-1)^2=0$$

Since $p$ and $q$ are real numbers, then LHS is sum of two squares, wich are non-negative, therefore:

$$\begin{cases} q^2 + p = 0 & \Longrightarrow & p = -q^2 \\ pq - 1 = 0 & \Longrightarrow & -q^3-1 = 0 \end{cases}$$

Therefore $q = -1$ what means that $p = -1$, wich contradicts that $p+q = 1$.

But the second equation holds:

$$p^3+q^3+3pq-1= (-1)^3 + (-1)^3 + 3(-1)(-1) -1 = 0$$

Answer 2

HINT:

$$(q^2+p)^2+(pq-1)^2=0$$

Also, $$a^3+b^3+c^3-3abc=(a+b+c)(\sum a^2-bc)$$

Here $a=p,b=q,c=-1$

See If $a,b,c \in R$ are distinct, then $-a^3-b^3-c^3+3abc \neq 0$.

Answer 3

Write the equations as $\,f(p,q) = 0\,$ and $\,g(p,q) = 0,\,$ respectively.

Then $\ {\rm mod}\ p\!+\!q\!-\!1\!:\,\ f \equiv 2(q^4\!-q^2\!+1)\,$ and $\, g\equiv 0$.

So if $\,p\!+\!q=1\,$ then $\,f = g \iff q^4\!-q^2\!+1=0\iff q^6=-1,\ q^2\ne -1$