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Given that - $$a=a_1 a_2$$ $$b=b_1 b_2$$ $$c=c_1 c_2$$ $$h=a_2 b_1 + b_2 a_1$$ $$g=a_1 c_2 + a_2 c_1$$ $$f=b_1 c_2 + b_2 c_1$$

Find the relationship between $a, b, c, f, g, h$

My Attempt: I could not see how I could exploit the symmetry of the equations to directly get an answer, so I tried to solve them as 6 simultaneous equations by - $$a_2=\frac{a}{a_1}$$ $$b_2=\frac{b}{b_1}$$ $$c_2=\frac{c}{c_1}$$ and put these values into the remaining 3 equations to obtain - $$a_1 b_1 h = a {b_1}^2 + b {a_1}^2$$ $$a_1 c_1 g = a {c_1}^2 + c {a_1}^2$$ $$c_1 b_1 f = c {b_1}^2 + b {c_1}^2$$ Then, I treated the last 2 equations as quadratics in $a_1$ and $b_1$; found their values using the quadratic formula and input those into the first of the last 3 equations above.

Then I found the value of $c_1$ using that and then the value of the remaining - $a_1, a_2, b_1, b_2, c_2$.

Then, when I finally input these values into any of the equations I got a tautology (which, as I now realize - too late - was doomed to happen from the very beginning, due to my approach).

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So, How should I go about finding the relation between $a, b, c, f, g, h$?

OR,

Equally - How do I eliminate $a_1, a_2, b_1, b_2, c_1, c_2$ from the equations?

I just need a hint on how I could exploit the symmetry of the equations.

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  • $\begingroup$ What does "Find the relationship between a,b,c,f,g,h" mean exactly? One could argue that the initial six equalities already gives a relationship among them. Are you looking for a polynomial expression in a,b,c,f,g,h (and not involving the subscripted variables) that equals 0? $\endgroup$ – Greg Martin May 19 '17 at 3:36
  • $\begingroup$ Hint: consider $af^2+bg^2+ch^2$. $\endgroup$ – Greg Martin May 19 '17 at 3:38
  • $\begingroup$ @GregMartin Yes, you're right I need a single expression in a, b, c, f, g, h without the subscripted variables. As an example of such an expression - $\frac{c^{2}}{2a}=\frac{f^{3}}{5g^{2}} +h tan(\frac{b}{a})$ The expression could include logs and trig functions (although I don't think they will be necessary) $\endgroup$ – Quantum Sphinx May 19 '17 at 3:39
  • $\begingroup$ @GregMartin Yes, that is the kind of expression I need $\endgroup$ – Quantum Sphinx May 19 '17 at 3:43
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Notice that $a,b,c,h,g,f$ are represented by staggered multiplication of $a_1,a_2,b_1,b_2,c_1,c_2$, and one way to align them is to multiply them together, and then we could rearrange how they are combined together. So symmetry is the key here.

$$h\cdot g \cdot f=(a_2 b_1 + b_2 a_1)(a_1 c_2 + a_2 c_1)(b_1 c_2 + b_2 c_1)$$ $$=a_2b_1a_1c_2b_1c_2 + a_2b_1a_1c_2b_2c_1+a_2b_1a_2c_1b_1c_2 + a_2b_1a_2c_1b_2c_1$$$$+b_2a_1a_1c_2b_1c_2+b_2a_1a_1c_2b_2c_1+b_2a_1a_2c_1b_1c_2+b_2a_1a_2c_1b_2c_1$$

$$=ab_1^2c_2^2+abc+a_2^2b_1^2c+a_2^2c_1^2b+a_1^2c_2^2b+a_1^2b_2^2c+abc+b_2^2c_1^2a$$ $$=2abc + a(b_1^2c_2^2+b_2^2c_1^2)+b(a_1^2c_2^2+a_2^2c_1^2) + c(a_2^2b_1^2+a_1^2b_2^2)$$ $$=2abc+a((b_1c_2 + b_2c_1)^2-2b_1c_2b_2c_1)+b((a_1c_2+a_2c_1)^2-2a_1c_2a_2c_1)+c((a_2b_1+a_1b_2)^2-2a_2b_1a_1b_2)$$ Thus $$hgf=2abc + a(f^2-2bc)+b(g^2-2ac)+c(h^2-2ab)$$ $$hgf +4abc -af^2-bg^2-ch^2=0$$

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