Given x+y+z=6, xy+yz+zx=11, and x3+y3+z3=36, find the values of x, y, z.

Step 1: Substitute the given values into the identity. The given system of equations is: x + y + z = 6 xy + yz + zx = 11 x^3 + y^3 + z^3 = 36 The provided identity is: x^3 + y^3 + z^3 - 3xyz = (x + y + z)[(x + y + z)^2 - 3(xy + yz + zx)] Substitute the known values into the identity: 36 - 3xyz = (6)[(6)^2 - 3(11)] Step 2: Solve for xyz. Simplify the equation from Step 1: 36 - 3xyz = 6[36 - 33] 36 - 3xyz = 6[3] 36 - 3xyz = 18 Subtract 36 from both sides: -3xyz = 18 - 36 -3xyz = -18 Divide by -3: xyz = (-18)/(-3) xyz = 6 Step 3: Form the cubic polynomial. For a cubic polynomial with roots x, y, z, the general form is: P(t) = t^3 - (x+y+z)t^2 + (xy+yz+zx)t - xyz = 0 Using the values we have: x+y+z = 6 xy+yz+zx = 11 xyz = 6 Substitute these values into the polynomial form: P(t) = t^3 - 6t^2 + 11t - 6 = 0 Step 4: Find the roots of the polynomial to determine x, y, z. We need to find the roots of t^3 - 6t^2 + 11t - 6 = 0. By testing integer divisors of the constant term -6 (i.e., ± 1, ± 2, ± 3, ± 6): For t=1: 1^3 - 6(1)^2 + 11(1) - 6 = 1 - 6 + 11 - 6 = 0. So t=1 is a root. Since t=1 is a root, (t-1) is a factor of the polynomial. We can perform polynomial division or synthetic division. Using synthetic division with root 1: ` 1 | 1 -6 11 -6 | 1 -5 6 ----------------- 1 -5 6 0 ` The quotient is t^2 - 5t + 6. Now, we solve the quadratic equation t^2 - 5t + 6 = 0. This factors as (t-2)(t-3) = 0. The roots are t=2 and t=3. Thus, the roots of the polynomial are 1, 2, 3. The real numbers x, y, z are the roots of this polynomial. Since the system is symmetric, the values of x, y, z are 1, 2, 3 in any order. The values for x, y, z are 1, 2, 3. The cubic polynomial is t^3 - 6t^2 + 11t - 6 = 0. Send me the next one 📸