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If $a,b\in\mathbb{R}$ and $\displaystyle \frac{a^3+4a}{3a^2+5}=-1.$ and $\displaystyle \frac{b^3+4b}{3b^2+5}=1$. Then $(a+b)^5=$

What I try : From the above data, we have

$\displaystyle a^3+3a^2+4a+5=0\cdots (1)$

$\displaystyle b^3-3b^2+4b-5=0\cdots (2)$

Adding both , We get

$\displaystyle a^3+b^3+3(a^2-b^2)+4(a+b)=0$

$(a+b)\bigg[a^2-ab+b^2+3a-3b+4\bigg]=0$

So either we get $a+b=0$ or

$a^2+b^2-ab+3a-3b+4=0$

But answer is $a+b=0\Longrightarrow (a+b)^5=0$

How can I prove that other factor is non zero,

Please have a look on that , thanks

jacky
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2 Answers2

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Observe that $$a^2-ab+b^2+3a-3b+4=\left(\frac{a+b}{2}\right)^2+3\left(1+\frac{a-b}{2}\right)^2+1$$

jjagmath
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First, we can prove that there is only one real value of $a$ that satisfies $\frac{a^3+4a}{3a^2+5} = -1$. We can see this from the cubic equation $$a^3 + 3a^2 + 4a + 5 = 0.$$ The derivative of the cubic polynomial is $3a^2 + 6a + 4 = 3(a+1)^2 + 1$, which is always positive. This means that the cubic polynomial is always increasing. After the first real root, the polynomial will always be positive and never hit $0$ again.

Second, for every value of $b$ such that $\frac{b^3 + 4b}{3b^2+5} = 1$, we have $\frac{(-b)^3 + 4(-b)}{3(-b)^2 + 5} = -1$. In other words, every solution to $b$'s equation is the negation of a solution to $a$'s equation. Since there was only one possibility for $a$, there can only be one possibility for $b$, and $b = -a$.

Misha Lavrov
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  • The statement "Every term of the cubic polynomial is (strictly) increasing in $a$" is wrong. $3a^2$ is not increasing. – jjagmath Mar 31 '24 at 19:16
  • "after the first real root, the polynomial will always be positive and never hit $0$ again." Nope. A cubic polynomial can have all its coefficients positive and still have $3$ real roots. – jjagmath Mar 31 '24 at 19:21
  • @jjagmath You are right, I got sloppy. I knew the polynomial had to be increasing, and I saw what looked like a reason for the polynomial to be increasing, and so I wrote it down without actually checking if it made sense. The argument is correct now. – Misha Lavrov Mar 31 '24 at 20:17