Let $a$ be a root of the cubic $x^3-21x+35=0$. Prove that $a^2+2a-14$ is a root of the cubic. 
Let $a$ be a root of the cubic $x^3-21x+35=0$. Prove that $a^2+2a-14$
  is a root of the cubic.

My effort
Working backwards I let $P(x)$ be a polynomial with roots $a,a^2+2a-14$ and $r$.
Thus, $$P(x)=(x-a)(x-r)(x-(a^2+2a-14))$$
Expanding, I get 
$$P(x) =(x^2-(a+r)x+ar)(x-(a^2+2a-14)) $$
$$P(x) =x^3+x^2[-(a^2+2a-14)-(a+r)]+x[(a+r)(a^2+2a-14)+ar]-ar(a^2+2a-14)$$
Equating coefficients of $P(x)$ with the given cubic $x^3-21x+35=0$ I have the following system of equations :
\begin{array}
\space (a^2+2a-14)+(a+r)&=0 \\ 
(a+r)(a^2+2a-14)+ar&=-21 \\
-ar(a^2+2a-14)&=35 \\
\end{array}
From the first equation I have $(a^2+2a-14) =-(a+r) $ which, substituted in the other two equations ,it yields 
\begin{array}
\space -(a+r)^2+ar &=-21 \\
ar(a+r) &=35 \\
\end{array}
Rearranging the second equation for $ar$ I have $ar=\cfrac{35}{(a+r)}$ which I now substitute into the first eq. to get:
\begin{array}
\space -(a+r)^2+\cfrac{35}{(a+r)}&=-21  \\
-(a+r)^3+35 +21(a+r) &=0 \\
\end{array}

My problem now is that the last equation looks pretty darn close to
  $x^3-21x+35=0$ but some signs are not in the right place,which makes
  me wonder if I have made some careless mistake(I have already checked
  but I don't see it) or if I have  left some algebraic manipulations to do.


Since the term in $x^2$ is missing, the sum of the three roots is zero; so $a^2+2a-14$ is a root if and only if $-a-(a^2+2a-14)=-a^2-3a+14$ is also a root.
Since
$$
(x-a^2-2a+14)(x+a^2+3a-14)=x^2+ax-a^4 - 5a^3 + 22a^2 + 70a - 196
$$
and the remainder of $-t^4 - 5t^3 + 22t^2 + 70t - 196$ divided by $t^3-21t+35$ is $t^2-21$, we have
$$
(x-a^2-2a+14)(x+a^2+3a-14)=x^2+ax+a^2-21
$$
so
\begin{align}
(x-a)(x-a^2-2a+14)(x+a^2+3a-14)
&=(x-a)(x^2+ax+a^2-21)\\
&=x^3-21x-a^3+21a\\
&=x^3-21x+35
\end{align}
is the required factorization.
