We need to solve x^3 - 7x + 6 for rational roots.

x^3 - 7x + 6 = 0

=> x^3 - x^2 + x^2 - x - 6x + 6 = 0

=> x^2(x - 1) + x(x - 1) - 6(x - 1) = 0

=> (x -1)(x^2 + x - 6) = 0

=> (x - 1)(x^2 + 3x - 2x - 6) = 0

=> (x - 1)(x(x + 3) - 2(x + 3)) = 0

=> (x - 1)(x - 2)(x + 3) = 0

=> x = 1, x = 2 and x = -3

**The rational roots of the polynomial are (1, 2, -3)**

We'll write 7x = 6x + x

We'll re-write the equation:

x^3 - 6x - x + 6 = 0

We'll find the roots of the equation between the divisor of the last terms divided by the divisors of the coefficient of the first term.

D6 = +/-1 ; +/-2 ; +/-3 ; +/-6

D1 = +/-1

We'll put x = 1:

1^3 - 6 - 1 + 6 = 0

0 = 0

So x = 1 is the first root of the equation.

Now, we'll use Viete's relations:

x1 + x2 + x3 = 0

1 + x2 + x3 = 0

x2 + x3 = -1

x1*x2 + x1*x3 + x2*x3 = -7

x2 + x3 + x2*x3 = -7

But

x2 + x3 = -1

-1 + x2*x3 = -7

x2*x3 = -7+1

x2*x3 = -6

x^2 + x - 6 = 0

We'll apply the quadratic formula:

x2 = [-1+sqrt(1+24)]/2

x2 = (-1+5)/6

x2 = 2/3

x3 = -6/6

x3 = -1

The roots of the equation are: {-1 ; 2/3 ; 1}.