If $a^4+b^4+a^2 b^2=273$ and $a^2+b^2-a b=21$, then one of the values of $\left(\frac{1}{a}+\frac{1}{b}\right)$ is:

$-\frac{3}{4}$

If $a^4+b^4+a^2 b^2=273$

$a^2+b^2-a b=21$,---- (A)

then one of the values of $\left(\frac{1}{a}+\frac{1}{b}\right)$= ?

We know that,

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

273 = 21 ($a^2+b^2+a b$)

($a^2+b^2+a b$) = 13---- (B)

From A and B equations ,

ab = -4

a² + b² = 17

( a + b )² = a² + b² + 2ab

( a + b )² = 17+ 2(-4)

( a + b )² =9

a + b  = 3

Now, $\left(\frac{1}{a}+\frac{1}{b}\right)$ = \(\frac{a + b}{ab}\) = -\(\frac{3}{4}\)