Show that solutions of $a^3+b^3=c^3+d^3$ and $a+b=c+d$ for distinct $a,b,c,d$ are non-existent.

Question

Question:

Show that there do not exist real, distinct $a,b,c,d$ such that $a^3+b^3=c^3+d^3$ and $a+b=c+d$.

My attempt:

$$a^3+b^3=c^3+d^3\implies(a+b)(a^2+b^2-ab)=(c+d)(c^2+d^2-cd)$$ Since $a+b=c+d$: $$\implies a^2+b^2-ab=c^2+d^2-cd$$ Substituting $b=c+d-a$ from second equation: $$\implies a^2+(c+d-a)^2=c^2+d^2-cd$$ $$\implies a^2+a^2+c^2+d^2+2cd-2ac-2ad=c^2+d^2-cd$$ $$\implies 2a^2+3cd-2ac-2ad=0$$

I am stuck here, and need some hints to proceed.

Answer 1

It's true if $(a+b)(c+d)\neq 0$ (if not, then $(a,b,c,d)=(a,-a,c,-c)$ is a counterexample for any $a,c\in\Bbb R_{\neq 0}$), so I assume it.

$a^3+b^3=c^3+d^3\iff (a+b)\left(a^2-ab+b^2\right)=(c+d)\left(c^2-cd+d^2\right)$. Since $(a+b)(c+d)\neq 0$, this is equivalent to

$$a^2-ab+b^2=c^2-cd+d^2\ \ \ (1)\ \ \iff (a+b)^2-3ab=(c+d)^2-3cd$$

But $a+b=c+d$, so $-3ab=-3cd\iff ab=cd$. Then $(1)$ implies $a^2+b^2=c^2+d^2$, so $a^2-c^2=d^2-b^2$, so $(a+c)(a-c)=(d+b)(d-b)$. But $a-c=d-b\neq 0$ (because $a+b=c+d$ and $a,b,c,d$ are distinct), so $a+c=d+b$. Adding this with $a+b=c+d$ gives $a=d$, contradiction.

Answer 2

$$x^3+y^3=(x+y)^3-3xy(x+y)$$

Set $x=a,y=b; x=c,y=d$ to find $ab=cd$

So if $a,b$ are the roots of $$t^2+pt+q=0$$

so will be $c,d$

Answer 3

Assuming $(a+b)\ne 0$, observe that,\begin{align}a^3+b^3=c^3+d^3&\implies(a+b)((a+b)^2-3ab))=(c+d)((c+d)^2-3cd))\\&\implies (a+b)((a+b)^2-3ab))=(a+b)((a+b)^2-3cd))\\&\implies ab=cd\\&\implies \dfrac{a}{c}=\dfrac{b}{d}=k\qquad(a,d,c,d\ \text{are}\ \color{crimson}{\text{distinct}})\end{align}Then $a=ck$ and $b=dk$. Now, $$a+b=c+d\implies k(c+d)=(c+d)\implies k=1$$