Question

Eliminate $x,y,z$ from the equations $\frac{x^2(y+z)}{a^3}=\frac{y^2(z+x)}{b^3}=\frac{z^2(x+y)}{c^3}=\frac{xyz}{abc}=1$.

Answer 1

$\frac{x^2(y+z)}{a^3}=\frac{y^2(z+x)}{b^3}=\frac{z^2(x+y)}{c^3}=\frac{xyz}{abc}=1$

$\Rightarrow x^2(y+z)=a^3,y^2(z+x)=b^3,z^2(x+y)=c^3$ and $xyz=abc$

Mutiplying all the terms

$\Rightarrow x^2(y+z)=a^3,y^2(z+x)=b^3,z^2(x+y)=c^3\Rightarrow x^2(y+z)y^2(z+x)z^2(x+y)=a^3{b}^3{c}^3\Rightarrow x^2{y}^2{z}^2(y+z)(z+x)(x+y)=a^3{b}^3{c}^3\Rightarrow a^2{b}^2{c}^2(y+z)(z+x)(x+y)=a^3{b}^3{c}^3\Rightarrow (y+z)(z+x)(x+y)=abc\Rightarrow (yz+xy+z^2+zx)(x+y)=abc\Rightarrow xyz+y^{2}z+x^{2}y+x{y}^2+z^{2}x+z^{2}y+x^{2}z+xyz=abc\Rightarrow x^2(y+z)+y^2(z+x)+z^2(x+y)+2xyz=abc\Rightarrow a^3+b^3+c^3+2abc=abc\Rightarrow a^3+b^3+c^3+abc=0$