Cassarrim1

2022-02-01

If k is even number then for what value of x and y,

${x}^{2k}-{\left(xy\right)}^{k}+{y}^{2k}=0$

nebajcioz

Beginner2022-02-02Added 15 answers

Suppose y is zero, then the only solution to ${x}^{2k}=0\text{}\text{is}\text{}x=0$ . Thus, $x=y=0$ is one of the possible solutions.

Now assume y is not zero. Then dividing everything by$y}^{2k$ you get

${\left(\frac{x}{y}\right)}^{2k}-{\left(\frac{x}{y}\right)}^{k}+1=0$

or, after substituting$z={\left(\frac{x}{y}\right)}^{k}$ ,

${z}^{2}-z+1=0$

This quadratic equation has no real solutions.

This means$x=y=0$ is the only possible solution, and even that for $k>0$ because when $k=0$ the expression $0}^{0$ is not well-defined.

Now assume y is not zero. Then dividing everything by

or, after substituting

This quadratic equation has no real solutions.

This means