Question

If x^2+y^2+z^2-xy-yz-zx=0, prove that x = y = z = 0​

Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

${x}^{2} + {y}^{2} + {z}^{2} - xy - yz - zx = 0$

Multiply both sides by 2, we get

$2{x}^{2} + 2{y}^{2} + 2{z}^{2} - 2xy - 2yz - 2zx = 0$

${x}^{2} + {x}^{2} + {y}^{2} + {y}^{2} + {z}^{2} + {z}^{2} - 2xy - 2yz - 2zx = 0$

can be re-arranged as

$( {x}^{2} + {y}^{2} - 2xy) + ( {y}^{2} + {z}^{2} - 2yz) + ( {z}^{2} + {x}^{2} - 2zx) = 0$

$\rm \: {(x - y)}^{2} + {(y - z)}^{2} + {(z - x)}^{2} = 0$

We know, Sum of squares is 0 only, if term itself is 0.

$\red{\rm\implies \:x - y = 0 \: \: \: and \: \: y - z = 0 \: \: and \: \: z - x = 0}$

$\red{\rm\implies \:x = y \: \: \: and \: \: y = z \: \: and \: \: z = x}$

$\red{\rm\implies \:x = y = z}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$