# 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.

• What's your question? It's rather uncomprehensible – enedil Feb 10 '19 at 16:39
• Given three expressions above I need to prove the fourth one. – questions about math Feb 10 '19 at 16:41
• Please come up with a more consise title. – Ramanujan Feb 10 '19 at 16:41
• But they are contradicting. – enedil Feb 10 '19 at 16:42
• @enedil can you elaborate? – questions about math Feb 10 '19 at 16:44

$$a^2x^2 + b^2y^2 + c^2z^2 = 0\tag1$$
$$a^2x^3 + b^2y^3 + c^2z^3 = 0\tag2$$
$$\frac 1x - a^2 = \frac 1y - b^2 = \frac 1z - c^2=k\tag3$$ $$(1)+(2)\implies$$ $$a^2(x^2+x^3)+ b^2(y^2+y^3)+ c^2(z^2+z^3) = 0$$ $$\iff$$ $$a^2x^3\left(1+\frac 1x \right) + b^2y^3\left(1+\frac 1y\right) + c^2z^3\left(1+\frac 1z\right) = 0$$ 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 $$a^4x^3 + b^4y^3 + c^4z^3 = 0.$$
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 $$a^4x^3+b^4y^3+c^4z^3 = (ax)a^3x^2 + (by)b^3y^2 + (cz)c^3z^2 = 0\cdot a^3x^2 + 0 \cdot b^3y^2 + 0 \cdot c^3z^2 = 0$$