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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.

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    $\begingroup$ What's your question? It's rather uncomprehensible $\endgroup$ – enedil Feb 10 '19 at 16:39
  • $\begingroup$ Given three expressions above I need to prove the fourth one. $\endgroup$ – questions about math Feb 10 '19 at 16:41
  • $\begingroup$ Please come up with a more consise title. $\endgroup$ – Ramanujan Feb 10 '19 at 16:41
  • $\begingroup$ But they are contradicting. $\endgroup$ – enedil Feb 10 '19 at 16:42
  • $\begingroup$ @enedil can you elaborate? $\endgroup$ – questions about math Feb 10 '19 at 16:44
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$$a^2x^2 + b^2y^2 + c^2z^2 = 0\tag1$$

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

$$\frac 1x - a^2 = \frac 1y - b^2 = \frac 1z - c^2=k\tag3$$ $(1)+(2)\implies$ $$a^2(x^2+x^3)+ b^2(y^2+y^3)+ c^2(z^2+z^3) = 0$$ $\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 $$a^4x^3 + b^4y^3 + c^4z^3 = 0. $$

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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).

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  • $\begingroup$ +1. Nice observation. I completely missed it. $\endgroup$ – Shivering Soldier Feb 10 '19 at 17:15

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