Question

Shew that $(a+b+c{)}^3-4(b+c\left)\right(c+a\left)\right(a+b)+5abc=0$ is the eliminant of $a{x}^2+b{y}^2+c{z}^2=ax+by+cz=yz+zx+xy=0$.

Answer 1

From the first two equations;-

$c(a{x}^2+b{y}^2)=-c^2{z}^2=-{(ax+by)}^2$

$\therefore a(a+c)x^2+2abxy+b(b+c)y^2=0⟶1$

From the third equation;-

$z=\frac{-xy}{x+y};$

Also

$cz=-(ax+by)$

So, $ax+by=\frac{cxy}{x+y}$

$\therefore x^2(a+b-c)xy+b{y}^2=0⟶2$

From $1$ & $2$ we obtain by

Cross multiplication

$=\frac{x^2}{b(b-c)(a-b-c)}=\frac{xy}{-ab(a-b)}=\frac{y^2}{a(a-c)(a-b+c)}$

Hence the eliminant is

$a^2{b}^2{(a-b)}^2=ab(a-c)(b-c)(a-b-c)(a-b+c);$

$ab{(a-b)}^2=(a-c)(b-c)(a-b-c)(a-b+c)$

$ab{(a-b)}^2=\{ab-(a+b)c+c^2\}\{{(a-b)}^2-c^2\}$

Crossing the term $ab{(a-b)}^2$ on each side and dividing by 'c', we have

$(a+b){(a+b)}^2-{(a-b)}^{2}c+abc-(a+b)c^2+c^2=0$

i.e., $a^2+b^2+c^2-b^{2}c-c^{2}a-b{c}^2-a^{2}c-b^{2}b-b^{2}a+3abc=0$

$\sum a^3={(a+b+c)}^3-3\sum a^{2}b-6abc$

$\sum a^{2}b=(b+c)(c+a)(a+b)-2abc$

Hence the eliminant is

$\{{(a+b+c)}^3-3(b+c)(c+a)(a+b)\}-\{(b+c)(c+a)(a+b)-2abc\}+3abc=0$

${(a+b+c)}^3-4(b+c)(c+a)(a+b)=5abc$