Just Minimize the Function!

The function $f$ will has the minimum value if $x^2 - x +1$ is at the minimum value.

Let $g(x) = x^2 - x + 1 \Rightarrow g(x) = \left( x - \frac{1}{2} \right)^2 + 1 - \frac{1}{4}$ .

$g(x) = \left( x - \frac{1}{2} \right)^2 + \frac{3}{4}$ .

We can clearly see that the minimum value of $g(x)$ occurs when $\left( x - \frac{1}{2} \right)^2$ is equal to zero; so that the minimum value of $g(x)$ is $\frac{3}{4}$ .

Now, we already have the minimum value of $x^2 - x + 1$ , so that the minimum value of function $f$ is

$\sqrt[3]{\frac{3}{4}} = \sqrt[a]{\frac{b}{c}}$

We have $a=3; \space b=3; \space c=4$ .

Hence, $a+b-c = 3+3-4 = \boxed{2}$ .

Let f(x) be y,

Therefore,y^3=x²-x+1

→x²-x+(1-y^3)=0

let the roots of equation be 'a' and 'b'

→a+b=1. And. ab=1-y^3

→y^3=1-ab (ab should have max value so that

y get min value)

ab has max value when a and b are same.

a=b=1/2

y=cube root of 3/4

3+3-4=2

{GRAPHICAL METHOD}..................x^2-x+1 >0 as D<0 ..IF WE DRAW THE GRAPH OF x^2-x+1 WHICH IS ALWAYS ABOVE X AXIS ,IT MINIMUM VALUE WILL BE THE VERTEX (Y COORDINATE)OF THE GRAPH WHICH IS A UPWARD PARABOLA ..SO MINIMUM VALUE OF x^2-x+1 IS -D/4a=3/4...so answer is cube root of 3/4 ....a =3,b=3,c=4...a=b-c=3+3-6=2