Question

Solve the following equations :
$x^2+xy+y^2=37,y^2+yz+z^2=61,z^2+zx+x^2=49.$

Answer 1

Multiplying the given equations by $x-y,y-z$ and $z-x$ respectively and adding.
$\sum (x^3-y^3)=37(x-y)=61(y-z)=49(z-x)=0$
$0=-12x+24y-12z$
or $2y=x+z,$ Hence $x,y,z$ are in $A.P$.
Hence the numbers can be taken as $y-d,y,y+d.$
Putting in the given equations, we get
$3y^2-3yd+d^2=37$
$3y^2+3yd+d^2=61$
$3y^2+d^2=49$
Putting the value of $3y^2+d^2$ in any, we get $yd=4$ and now solve this with $3y^2+d^2=49$.
or $3y^2+\frac{16}{y^2}=49$
or $3y^4-49y^2+16=0$
$(y^2-16)(3y^2-1)=0$
$\therefore y^2=16$ or $y=4,-4$
$\therefore d=1,-1$ $\because yd=4$
$x=y-d=3,-3;$ $z=y+d=5,-5$
Hence the numbers are $(3,4,5)$ or $(-3,-4,-5)$
Note: $3y^2-1=0$ $\Rightarrow y^2=1/3$ and these values do not satisfy the given relations.