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

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

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

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

$$(\because (a + b)^2 = a^2 + 2ab + b^2)$$

= $$(x^2 + y^2)^2 = 21 + x^2y^2$$ ----(1)

$$x^2 + xy + y^2 = 7$$

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

From eq(1) and (2),

$$(7 - xy)^2 = 21 + x^2y^2$$

$$49 + x^2y^2 - 14xy = 21 + x^2y^2$$

14xy = 49 - 21

xy = 28/14 = 2

From eq(2),

$$x^2 + y^2 = 7 - 2$$

$$x^2 + y^2 = 5$$

Now,

$$\left(\frac{1}{x^2} + \frac{1}{y^2}\right)$$

= $$\frac{x^2 + y^2}{x^2y^2}$$

= $$\frac{5}{2^2}$$ 

= $$\frac{5}{4}$$