Precalculus algebra exercise Hi I need to solve this problem and I don’t know how so I’d appreciate a hint.
If 
$a^2x^2 + b^2y^2 + c^2z^2 = 0$
$a^2x^3 + b^2y^3 + c^2z^3 = 0$
$\frac 1x - a^2 = \frac 1y - b^2 = \frac 1z - c^2$
Then $a^4x^3 + b^4y^3 + c^4z^3 = 0$
I think that  $a^4x^3 + b^4y^3 + c^4z^3 = 0$ is a factor in an expression which can be found by manipulating the three given equations. I can see that
$\frac 1x - a^2 - \frac 1y + b^2 = 0$
So I tried to add, subtract, multiply given equations.
 A: Suppose that
$$
\begin{cases}
(ax)^2 + (by)^2 + (cz)^2 = 0\\
a^2x^3 + b^2y^3 + c^2z^3 = 0\\
\frac 1x - a^2 = \frac 1y - b^2 = \frac 1z - c^2
\end{cases}
$$
However, for any $a, x \in \mathbb R$, $(ax)^2 \geq 0$. So we have
$$
0 = (ax)^2 + (by)^2 + (cz)^2 \geq 0 + 0 + 0 = 0
$$
Therefore in each inequality, there's equality, so
$$
\begin{cases}
ax = 0\\
by = 0\\
cz = 0
\end{cases}
$$
This implies
$$
a^4x^3+b^4y^3+c^4z^3 = (ax)a^3x^2 + (by)b^3y^2 + (cz)c^3z^2 = 0\cdot a^3x^2 + 0 \cdot b^3y^2 + 0 \cdot c^3z^2 = 0
$$
How funny that we didn't need to use equations (2) and (3).
A: $$a^2x^2 + b^2y^2 + c^2z^2 = 0\tag1$$
$$a^2x^3 + b^2y^3 + c^2z^3 = 0\tag2$$
$$\frac 1x - a^2 = \frac 1y - b^2 = \frac 1z - c^2=k\tag3$$
$(1)+(2)\implies$
$$a^2(x^2+x^3)+ b^2(y^2+y^3)+ c^2(z^2+z^3) = 0$$
$\iff $
$$a^2x^3\left(1+\frac 1x \right) + b^2y^3\left(1+\frac 1y\right) + c^2z^3\left(1+\frac 1z\right) = 0$$
Now using $(3) $, we can write, $$a^2x^3(1+k+a^2 ) + b^2y^3(1+k+b^2) + c^2z^3(1+k+c^2) = 0.$$
$$(a^2x^3 + b^2y^3 + c^2z^3)+k (a^2x^3 + b^2y^3 + c^2z^3)+a^4x^3 + b^4y^3 + c^4z^3 = 0.$$
Using  $(2) $, we can conclude that 
$$a^4x^3 + b^4y^3 + c^4z^3 = 0. $$
