If $x^4+x^2 y^2+y^4=21$ and $x^2+x y+y^2=7$, then the value of $\left(\frac{1}{x^2}+\frac{1}{y^2}\right)$ is:

$\frac{5}{4}$

We know that,

x⁴ + x²y² + y⁴ = (x² – xy + y²) (x² + xy + y²)

If $x^4+x^2 y^2+y^4=21$

$x^2+x y+y^2=7$------(1)

Then the value of $\left(\frac{1}{x^2}+\frac{1}{y^2}\right)$

$x^2-x y+y^2=\frac{21}{7}$ = 3----(2)

Now, $\left(\frac{1}{x^2}+\frac{1}{y^2}\right)$ can be written as \(\frac{x^2 + y^2}{x^2y^2}\)---(*)

So the value of $x^2+y^2$ from equations 1 and 2 = 5

and $x^2y^2$ = 4

Put these values in (*) = \(\frac{5}{4}\)