243. Nice ! (a+b) is constant.

by Virgil Nicula, Mar 4, 2011, 8:29 PM

PP. Prove that the sum of two real numbers $ a$ , $ b$ which verifiy the relation $ a^3+b^3+3\left(a^2+b^2\right)+4(a+b)+4=0\ (*)$ is constant.

Method 1. Let $ a$ , $ b$ be two real numbers which verify the relation $(*)$ . Exists a real number $ p$ so that $ a^3+3a^2+4a-p=0$ and $ b^3+3b^2+4b+p+4=0$ .

Consider the function $ f(x)=x^3+3x^2+4x+p$ , where $ x\in \mathbb R$ . Observe that $ f(a)=2p$ and $ f(b)=-4$ . Prove easily that the function $ f$ is strict increasing

(without derivatives !) and the graph of the function $ f$ has symmetry point $ S(-1,p-2)$ , i.e. $ f(-2-x)=2(p-2)-f(x)$ . For $ x: =a$ obtain that

$ f(-2-a)=2(p-2)-f(a)=2p-4-2p=-4=f(b)$ , i.e. $ f(-2-a)=f(b)$ . Since the function $ f$ is injective (it is strict increasing) obtain that

$ -2-a=b$ , i.e. $ a+b=-2$ (constant).

Method 2. Let $ a$ , $ b$ be two real numbers which verify the relation $(*)$ . If $ s=a+b$ si $ p=ab$ , then relation $(*)$ becomes $ s^3+3s^2+4s+4-3(s+2)p=0$ .

If their sum is constant, then the equation in $ s$ has independent roots by $ p$ , i.e. the equations $ \left\|\ \begin{array}{c} s^3+3s^2+4s+4=0\\\ -3(s+2)=0\end{array}\ \right\|$ have at least a common real root,

i.e. $ s=-2$ $\implies$ $ a+b=-2$ .

Another example. $\{a,b\}\subset\mathbb R\ \ \wedge\ \ a^3+b^3+a+b+2=3\left(a^2+b^2\right)\ \implies\ a+b=2$ .
This post has been edited 11 times. Last edited by Virgil Nicula, Nov 22, 2015, 12:45 PM

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