Find a and b, that there are two roots of $x^3 - 5x^2+ 7x = a$, that they are roots of $x^3 - 8x + b = 0$
I find that $a+b=5x(-x+3) $ Also I tried to solve second equation to get roots, depending on b, but my approach was unsuccessful.
Any help is appreciated!
On
Let $P(x) = x^3-5x^2+7x-a$ and $Q(x) = x^3-8x+b$ and $\gamma, \delta$ be their common roots.
We can decompose them as
$$\begin{cases} P(x) &= (x-\alpha)M(x),\\ Q(x) &= (x-\beta)M(x) \end{cases}\quad\text{ where }\quad M(x) = (x-\gamma)(x-\delta) $$ Subtract them, we get
$$5x^2-15x+b+a = Q(x)-P(x) = (\alpha-\beta)M(x)$$ By comparing the coefficients of $x^2$, we find $\alpha - \beta = 5$ and $$M(x) = x^2 - 3x + \frac{a+b}{5}$$
Notice
$$(2-\alpha)M(x) = P(x) - (x-2)M(x) = \frac15((b+a-5)x + (3a-2b))$$ If $\alpha \ne 2$, the LHS is a polynomial of degree $2$ while the RHS is a polynomial of degree at most $1$. This is impossible, so $\alpha = 2$ and
$$(b+a-5)x + (3a-2b) = 0 \quad\implies\quad \begin{cases} b+a-5 &= 0\\ 3a-2b &= 0 \end{cases} \quad\implies\quad (a,b) = (2,3) $$
They are roots of the equation $$x^2=3x-\frac{a+b}{5},$$ which you got.
Thus, since $x^3-8x+b=0$ has these two roots, we see that the equation $$x\left(3x-\frac{a+b}{5}\right)-8x+b=0$$ or $$x^2-\frac{1}{3}\left(\frac{a+b}{5}+8\right)x+\frac{b}{3}=0$$ has these two roots.
Thus, we have the following system $$\frac{1}{3}\left(\frac{a+b}{5}+8\right)=3$$ and $$\frac{b}{3}=\frac{a+b}{5},$$ which gives $a=2$ and $b=3$.