Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$x^2-yz=a^2$

$y^2-zx=b^2$

$z^2-xy=c^2$

How to solve this equation for $x,y,z$. Use elementary methods to solve (elimination, substitution etc.).

Given answer is:$x=\pm\dfrac{a^4-b^2c^2}{\sqrt {a^6+b^6+c^6-3a^2b^2c^2}}\,$, $y=\pm\dfrac{b^4-a^2c^2}{\sqrt {a^6+b^6+c^6-3a^2b^2c^2}}\,$ ,$z=\pm\dfrac{c^4-b^2a^2}{\sqrt {a^6+b^6+c^6-3a^2b^2c^2}}$

share|improve this question

2 Answers 2

Observe that $$(a^2)^2-b^2\cdot c^2=(x^2-yz)^2-(y^2-zx)(z^2-xy)=x(x^3+y^3+z^3-3xyz)$$

Similarly, $$b^4-c^2a^2=y(x^3+y^3+z^3-3xyz)\text{ and }c^4-a^2b^2=z(x^3+y^3+z^3-3xyz)$$

So, $$\frac x{a^4-b^2c^2}=\frac y{b^4-c^2a^2}=\frac z{c^4-a^2b^2}=\frac1{x^3+y^3+z^3-3xyz}=k\text{(say)}$$

Now, $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)=(x+y+z)\frac{\{(x-y)^2+(y-z)^2+(z-x)^2\}}2$

$x+y+z=k(a^4-b^2c^2+b^4-c^2a^2+c^4-a^2b^2)=\frac k2\{(a^2-bc)^2+(b^2-ca)^2+(c^2-ab)^2\}$

and $x-y=k\{a^4-b^2c^2-(b^4-c^2a^2)\}=k(a^2+b^2+c^2)(a^2-b^2)$

$\implies (x-y)^2+(y-z)^2+(z-x)^2=k^2(a^2+b^2+c^2)^2\{(a^2-b^2)^2+(b^2-c^2)^2+(c^2-a^2)^2\}=k^2(a^2+b^2+c^2)^2\cdot 2(a^4-b^2c^2+b^4-c^2a^2+c^4-a^2b^2)$

$\implies x^3+y^3+z^3-3xyz=k^3\{(a^2+b^2+c^2)(a^4-b^2c^2+b^4-c^2a^2+c^4-a^2b^2)\}^2=k^3(a^6+b^6+c^6-3a^2b^2c^2)^2 $

$$\implies\frac1k=x^3+y^3+z^3-3xyz=k^3(a^6+b^6+c^6-3a^2b^2c^2)^2$$

$$\implies k^2=\frac1{a^6+b^6+c^6-3a^2b^2c^2}$$ as $a^6+b^6+c^6-3a^2b^2c^2=(a^2+b^2+c^2)\frac{\{(bc-a^2)^2+(ca-b^2)^2+(ab-c^2)^2\}}2\ge 0$ for real $a,b,c$

share|improve this answer

Subtract them we get $$(x^2-y^2)+xz-yz = a^2-b^2 \implies (x-y)(x+y+z) = (a-b)(a+b)$$ $$(y^2-z^2)+yx-zx = b^2-c^2 \implies (y-z)(x+y+z) = (b-c)(b+c)$$ $$(z^2-x^2)+zy-xy = c^2-a^2 \implies (z-x)(x+y+z) = (c-a)(c+a)$$ Now consider the cases $x+y+z=0$ and $x+y+z\neq 0$. I trust you can finish it off from here.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.