A number related to the roots of a quartic polynomial is a root of a cubic polynomial

So here is the problem, $a$ and $b$ are two distinct real roots of $f(x)=0$ where $f(x)=x^4-6x+3$, show that $(a+b)^2$ is a root of $g(x)=x^3-12x-36$.

I have tried many methods, such as substitution, expanding the polynomial, changing it to different form, and reduce the power of $x$, but still could not make any process.

Can anyone help me for some suggestion?

Thank you very much.

• It should be “$a$ and $b$ are the two distinct roots of $f(x)=0$”. – egreg Apr 12 '14 at 22:14
• I just asked Wolframalpha the roots and checked with a calculator; with those values I get $g((a+b)^2)\approx-0.008$. Taking $2a$ or $2b$ I get nothing near $0$, so the hypothesis that they are the two distinct roots seems essential. – egreg Apr 12 '14 at 22:20
• Yes, I meant to say that the conjecture seems indeed valid. – egreg Apr 12 '14 at 22:34

We have $$a^{4}-6a+3=0$$ and $$b^{4}-6b+3=0.$$ Subtracting the second from the first and cancelling the factor $a-b$, we obtain, $$(a+b)\{(a+b)^{2}-2ab\}=6.$$ Squaring this, and denoting $a+b$ and $ab$ by $t$ and $x$ respectively, we have, $$t^{6}-4t^{2}(xt^{2}-x^{2})-36=0.$$ So we must show that $xt^{2}-x^{2}=3$. Now, for that we denote by $c$ and $d$, the other roots of ${\rm f}(x)$. We have $$abc+abd+acd+bcd=6.$$ Also, using $a+b+c+d=0$ and $abcd=3$, we obtain $$t(\frac {3}{x}-x)=6.$$ But, we had found before that $t(t^{2}-2x)=6$ and hence, we have(since $t \neq 0$), $$t^{2}-2x=\frac {6}{t}=\frac{3}{x}-x.$$ Thus, $$xt^{2}-x^{2}=3.$$