Show that $2Q(x)$ can be written as the sum of three perfect squares (a) Show that $x^3-3xbc+b^3+c^3$ can be written in the form $(x+b+c)Q(x)$, where $Q(x)$ is a qudaratic expression.
(b) Show that $2Q(x)$ can be written as the sum of three expressions, each of which is a perfect square.
My attempt at solution: (a) Through long division, I arrived at $Q(x) = x^2-(b+c)x+b^2-bc+c^2$
(b) The only way that I know to rewrite a quadratic expression as perfect squares is through completing the square. So $$(x-\frac{b+c}{2})^2-(\frac{b+c}{2})^2+b^2-bc+c^2$$ $\therefore (x-\frac{b+c}{2})^2 + (b-c)^2 - \frac{1}{4}(b-c)^2$
$\therefore 2Q(x) = 2(x-\frac{b+c}{2})^2 + (b-c)^2 + \frac{1}{2}(b-c)^2$
This is the best that I've managed to come up with, but it doesn't seem to be right as the question mentioned it can be written as the sum of three perfect squares.
 A: I will put the Hessian matrix of $x^2 + y^2 + z^2 - yz - zx -xy$   and display , with square matrices $PQ = QP = I,$ $P^T HP = D$  is diagonal, so $Q^T D Q = H$
I will type in the final outcome first. The $Q^T DQ$ version reads
$$  \left( x - \frac{y}{2}  - \frac{z}{2} \right)^2 + \frac{3}{4} (y-z)^2 =  x^2 + y^2 + z^2 - yz - zx - xy  $$
which does show that the quadratic form is rank two, not full rank. Furthermore, the form is zero  when $x=y=z$
and only then. This does suggest that we might try to write the form using $(u-v)^2$  and $(v-w)^2.$   It will work out better if $(w-u)^2$ is included somehow
$$  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc     \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc $$
$$ P^T H P = D  $$
$$\left( 
\begin{array}{rrr} 
1 & 0 & 0 \\ 
 \frac{ 1 }{ 2 }  & 1 & 0 \\ 
1 & 1 & 1 \\ 
\end{array}
\right) 
\left( 
\begin{array}{rrr} 
2 &  - 1 &  - 1 \\ 
 - 1 & 2 &  - 1 \\ 
 - 1 &  - 1 & 2 \\ 
\end{array}
\right) 
\left( 
\begin{array}{rrr} 
1 &  \frac{ 1 }{ 2 }  & 1 \\ 
0 & 1 & 1 \\ 
0 & 0 & 1 \\ 
\end{array}
\right) 
 = \left( 
\begin{array}{rrr} 
2 & 0 & 0 \\ 
0 &  \frac{ 3 }{ 2 }  & 0 \\ 
0 & 0 & 0 \\ 
\end{array}
\right) 
$$
$$  $$
$$ Q^T D Q = H  $$
$$\left( 
\begin{array}{rrr} 
1 & 0 & 0 \\ 
 -  \frac{ 1 }{ 2 }  & 1 & 0 \\ 
 -  \frac{ 1 }{ 2 }  &  - 1 & 1 \\ 
\end{array}
\right) 
\left( 
\begin{array}{rrr} 
2 & 0 & 0 \\ 
0 &  \frac{ 3 }{ 2 }  & 0 \\ 
0 & 0 & 0 \\ 
\end{array}
\right) 
\left( 
\begin{array}{rrr} 
1 &  -  \frac{ 1 }{ 2 }  &  -  \frac{ 1 }{ 2 }  \\ 
0 & 1 &  - 1 \\ 
0 & 0 & 1 \\ 
\end{array}
\right) 
 = \left( 
\begin{array}{rrr} 
2 &  - 1 &  - 1 \\ 
 - 1 & 2 &  - 1 \\ 
 - 1 &  - 1 & 2 \\ 
\end{array}
\right) 
$$
$$  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc     \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc $$
