mjhjmtu

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.

$$a^2x^2 + b^2y^2 + c^2z^2 = 0tag1$$

$$a^2x^3 + b^2y^3 + c^2z^3 = 0tag2$$

$$frac 1x - a^2 = frac 1y - b^2 = frac 1z - c^2=ktag3$$
$(1)+(2)implies$
$$a^2(x^2+x^3)+ b^2(y^2+y^3)+ c^2(z^2+z^3) = 0$$
$iff $
$$a^2x^3left(1+frac 1x right) + b^2y^3left(1+frac 1yright) + c^2z^3left(1+frac 1zright) = 0$$
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
$$a^4x^3 + b^4y^3 + c^4z^3 = 0. $$

Suppose that
$$
begincases
(ax)^2 + (by)^2 + (cz)^2 = 0\
a^2x^3 + b^2y^3 + c^2z^3 = 0\
frac 1x - a^2 = frac 1y - b^2 = frac 1z - c^2
endcases
$$
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
$$
begincases
ax = 0\
by = 0\
cz = 0
endcases
$$
This implies
$$
a^4x^3+b^4y^3+c^4z^3 = (ax)a^3x^2 + (by)b^3y^2 + (cz)c^3z^2 = 0cdot a^3x^2 + 0 cdot b^3y^2 + 0 cdot c^3z^2 = 0
$$

How funny that we didn't need to use equations (2) and (3).