Finding out the number of solutions If $a,b,c$ are three distinct real numbers for which there exists $x,y,z$ such that $$x+y+z=1$$$$ax+by+cz=t$$$$a^2x+b^2y+c^2z=t^2$$then $a^3x+b^3y+c^3z-t^3=P(t)$,be a polynomial in $t$,then the question is to find out the number of solutions to P(t)=0 .

My attempt at the solution

We are given three equations.From there we can solve for $x,y,z$ using Cramer's rule.by solving I got $$x=\frac{(t-b)(b-c)(c-t)}{(a-b)(b-c)(c-a)}$$  $$y=\frac{(a-t)(c-a)(t-c)}{(a-b)(b-c)(c-a)}$$    $$z=\frac{(a-b)(b-t)(t-a)}{(a-b)(b-c)(c-a)}$$  however plugging it in the equation I got $$P(t)=a^3( t-b)(b-c)(c-t)+b^3 (a-t)(c-a)(t-c)+c^3 (a-b)(b-t)(t-a)-t^3(a-b)(b-c)(c-a)=0$$ from here I couldn't proceed.please help me in this regard.thanks.
you are given $a^{3}x+b^{3}y+c^{3}z−t^{3}=P(t)$.Equate $P(t)=0$.Then multiply $(a+b+c)$ on both sides and then simplify to get
$t^{3}-(a+b+c)t^{2}+(ab+bc+ca)t-abc=0 $ which shows that the roots are $a,b,c$.
Now just simply observe that $P(a)=0$ and $P(b)=0$ and $P(c)=0$

Well just plug in $x,y,z$ respectively...

So when you calculate $P(a)$, you will find that $y,z$ simply vanish and $x=1$.
Hence it is a root...