Question

Solve the following equation :
$3x-4y+7z=0,2x-y-2z=0,3x^3-y^3+z^3=18.$

Answer 1

$3x-4y+7z=\mathrm{0.....}\left(i\right)2x-y-2z=\mathrm{0......}\left(ii\right)3x^3-y^3-z^3=\mathrm{8....}\left(iii\right)$

Applying cross multiplication on $\left(i\right)$ and $\left(ii\right)$

$\frac{x}{8+7}=\frac{-y}{-6-14}=\frac{z}{-3+8}\frac{x}{15}=\frac{y}{20}=\frac{z}{5}\frac{x}{3}=\frac{y}{4}=z=k\Rightarrow x=3k,y=4k,z=k$

Substituting $x,y$ and $z$ in $\left(iii\right)$

$\Rightarrow x=3k,y=4k,z=k3{\left(3k\right)}^3-{\left(4k\right)}^3{+\left(k\right)}^3=8k^3=1k=1\Rightarrow x=3,y=4,z=1$