$ \displaystyle x$ = 1, 2, â€“3

**Explanation**:

We are asked to solve the equation *x*Â³ â€“ 7x + 6 = 0, which is a cubic equation. We will find the roots using the trial-and-error method followed by factorization.

**1. Check for a Root via Substitution:** We can start by trying simple integer values of *x* to check for roots. Try *x* = 1:

*f *(1) = 1Â³ â€“ 7(1) + 6 = 1 â€“ 7 + 6 = 0

Therefore, *x* = 1 is a root of the polynomial.

**2. Perform Synthetic Division:** Now, we divide *x*Â³ â€“ 7*x *+ 6 by *x*Â â€“ 1 using synthetic division:

The quotient is *x*Â² + *x*Â â€“ 6, and the remainder is 0.

**3. Factor the Resulting Quadratic:** Now, factor the quadratic *x*Â² + *x â€“* 6. We look for two numbers that multiply to â€“6 and add up to 1. These numbers are 3 and â€“2.

So, the quadratic factors as:

*Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â x*Â² + *x* â€“ 6 = (*x â€“* 2)(*x* + 3)

**4. Final Factorization:** Therefore, the factorization of the cubic equation is:

(*x* â€“ 1)(*x â€“* 2)(*x* + 3) = 0

The solutions are:

*x* = 1, *x = *2, *x* = â€“3