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:
Answer & explanation
Correct answer: option 2
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, $\left(\frac{1}{a}+\frac{1}{b}\right)$ = \(\frac{a + b}{ab}\) = -\(\frac{3}{4}\)