Question

Shew that $b^3{c}^3+c^3{a}^3+a^3{b}^3=5a^2{b}^2{c}^2$ is the eliminant of $ax+yz=bc,by+zx=ca,cz+xy=ab,xyz=abc$.

Answer 1

Substituting $z=\frac{abc}{xy}$ in first three equations,

we have,

$a{x}^2-bcx+abc=0$

$b{y}^2-cay+abc=0$

$x^2{y}^2-abxy+ab{c}^2=0$

From the last two equations we have

$\frac{y^2}{a^{2}bc(c^2-bx)}=\frac{-y}{abc(x^2-bc)}-\frac{1}{ax(b^2-cx)}$

Hence, $a^{3}bcx(c^2-bx)(b^2-cx)=a^2{b}^2{c}^2{(x^2-bc)}^2$

i.e., $bc{x}^4-abc{x}^3+x^2(a{b}^3+a{c}^3-2b^2{c}^2)-a{b}^2{c}^{2}x+b^2{c}^2=0$

It remains to eliminate x between this equation and $a{x}^3-bcx+abc=0$

Multiply the first equation by 'a' and the equation $a{x}^2-bcx+abc=0$ by $bc(bc-a^2)x$ and subtract, then

$(a^3{b}^3+a^3{c}^3+b^3{c}^3-4a^2{b}^2{c}^2)x^2-a{b}^3{c}^{3}x+a^2{b}^3{c}^3=0$

Multiply $a{x}^2-bcx+abc=0$ by

$a{b}^2{c}^2$ and subtract from this last equation;-

$(a^3{b}^3+b^3{c}^3+c^3{a}^3-5a^2{b}^2{c}^2)x^2=0$

$\therefore a^3{b}^3+b^3{c}^3+c^3{a}^3=5a^2{b}^2{c}^2$