Showing $(a+b+c)(x+y+z)=ax+by+cz$ given other facts

Question

$$x^2-yz/a=y^2-zx/b=z^2-xy/c$$

None of these fractions are equal to 0.We need to show that,

$(a+b+c)(x+y+z)=ax+by+cz$

This question comes from a chapter that wholly deals with factoring homogeneous cyclic polynomials.I multiplied the three sides of the first equality by $abc$ but that yields an unfactorizable polynomial.I haven't had much luck in manipulating the first equality.So I tried to understand what I was trying to prove by expanding the second equality.We are trying to prove that,

$$a(y+z)+b(x+z)+c(x+y)=0$$

But the LHS is still unfactorizable.I tried to manipulate the first equality more,but they yielded nothing.However,I did find the following equality:

$$bc(x^2-yz)-ac(y^2-zx)+ab(z^2-xy)=a(y+z)+b(x+z)+c(x+y) $$

I would appreciate a very small hint.

score 3 · Accepted Answer · edited Apr 13 '17 at 12:21

3

Using Addendo formula $\displaystyle \frac Aa=\frac Bb=\cdots=\frac{A+B+\cdots}{a+b+\cdots},$

$$\frac{x^2-yz}a=\frac{y^2-zx}b=\frac{z^2-xy}c=\frac{x^2-yz+y^2-zx+z^2-xy}{a+b+c}\ \ \ \ (1)$$

Multiplying the numerator & the denominator of the first term by $x,$ and those of second by $y$ and those of third by $z$ $$\implies\frac{x^3-xyz}{ax}=\frac{y^3-xyz}{by}=\frac{z^3-xyz}{cz}=\frac{x^3+y^3+z^3-3xyz}{ax+by+cz}\ \ \ \ (2)\text{ (again applying Addendo)}$$

Equate $(1),(2)$ using Factorize the polynomial $x^3+y^3+z^3-3xyz$

edited Apr 13 '17 at 12:21

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answered Nov 30 '13 at 15:27

lab bhattacharjee

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Equating them gives $1/(a+b+c)=(x+y+z)/(ax+by+cz)$ and inversing both sides and multiplying both sides by $x+y+z$ gives us the required result.Thank you! – rah4927 Nov 30 '13 at 15:44
@rahul, my pleasure. Please feel free to get back to me if you have any query, concern – lab bhattacharjee Nov 30 '13 at 15:46
@rahul, nice to see that you are going through meticulously – lab bhattacharjee Nov 30 '13 at 15:49

Showing $(a+b+c)(x+y+z)=ax+by+cz$ given other facts

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