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.
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, $$xt^{2}-x^{2}=3.$$