if the root of the equation 3x^2-6x+1=0 are a and b,find the value of
[a^2+1/(b^2)][b^2+1/(a^2)]
First of all, I can't tell if this is
$\displaystyle \left(\frac{a^2 + 1}{b^2}\right)\left(\frac{b^2 + 1}{a^2}\right)$ or $\displaystyle \left(a^2 + \frac{1}{b^2}\right)\left(b^2 + \frac{1}{a^2}\right)$.
Anyway, to begin to answer your questions...
$\displaystyle 3x^2 - 6x + 1 = 0$
$\displaystyle x = \frac{6 \pm \sqrt{(-6)^2 - 4(3)(1)}}{2(3)}$
$\displaystyle = \frac{6 \pm \sqrt{36 - 12}}{6}$
$\displaystyle = \frac{6 \pm \sqrt{24}}{6}$
$\displaystyle = \frac{6 \pm 2\sqrt{6}}{6}$
$\displaystyle = \frac{3 \pm \sqrt{6}}{3}$.
So $\displaystyle a = \frac{3 - \sqrt{6}}{3}$ and $\displaystyle b = \frac{3 + \sqrt{6}}{3}$. Substitute these values into your expression and simplify.