Question

Solve the following equations:
$x^3+y^3+z^3=a^3,x^2+y^2+z^2=a^2,x+y+z=a$.

• A. $x=(a,0,0);y=(0,a,0);z=(0,0,a)$
• B. $x=(2a,a,0);y=(0,a,0);z=(0,a,a)$
• C. $x=(-a,0,2a);y=(a,a,0);z=(0,0,a)$
• D. None of the above

Answer 1

The correct option is A $x=(a,0,0);y=(0,a,0);z=(0,0,a)$

$x^3+y^3+z^3=a^3$ ......(i)

$x^2+y^2+z^2=a^2$ .....(ii)

$x+y+z=a$ .........(iii)

${(x+y+z)}^2=x^2+y^2+z^2+2xy+2yz+2zx{a}^2=a^2+2xy+2yz+2zx\Rightarrow xy+yz+zx=0$ ........(iv)

$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)a^3-3xyz=a(a^2-0)\Rightarrow xyz=0\Rightarrow x=0,y=0,z=0$

When $x=0$

$z^2+y^2=a^{2}z+y=a$

Solving both the sides, we get

$\Rightarrow z=a,0;y=0,a$

When $y=0$

$x^2+z^2=a^{2}y+z=a\Rightarrow x=a,0;z=0,a$

When $z=0$

$x^2+y^2=a^{2}x+y=a\Rightarrow x=0,a;y=a,0$

So, complete solution is $x=(a,0,0);y=(0,a,0);z=(0,0,a)$.