(2 + √3) and (1 + i) are roots of equation.

x⁶ - 6x⁵ + 10x⁴ -9x² + 6x - 2 = 0-------(i)

Let p(x) = x⁶ - 6x⁵ + 10x⁴ - 9x² + 6x - 2

\(\because\) (2 + √3) and (1 + i) are roots of equation (i)

\(\therefore\) (2 + √3) and (1 - i) are roots of equation (i)

⇒ (x - (2 + √3))(x - (2 - √3)) is a factor of p(x)

⇒ ((x - 2) - √3) ((x - 2) + √3) is a factor of p(x)

⇒ ((x - 2)² - 3) is a factor of p(x)

⇒ (x²- 4x + 1) is a factor of p(x)

Now,

\(\frac{P(x)}{x^2-4x + 1}=\frac{x^6-6x^5+10x^4-9x^2+6x-2}{x^2-4x+1}\) 

=(x⁴- 2x³ + x² + 6x + 14) (x²- 4x + 1) + \(\frac{56x-16}{x^2-4x+1}\)

\(\therefore\) x² - 4x + 1 is not a divisior of p(x).

\(\therefore\) (2 +√3) is not a root or given equation.

\(\therefore\) (1 + i) is a root of equation (i)

\(\therefore\) (x - (1 + i))(x - (1 - i)) is a factor of P(x)

⇒ ((x - 1)² - i²) is a factor of P(x)

⇒ (x²- 2x + 2) is a factor of P(x)

Now, 

\(\frac{P(x)}{x^2-2x + 2}\) = \(\frac{x^6-6x^5+10x^4-3x^2+6x-2}{x^2-2x+2}\)

= x⁴ - 4x³ + 8x + 7 + \(\frac{4x-16}{x^2-2x+2}\) 

Hence, x² - 2x + 2 is not a divisior of P(x).

\(\therefore\) (1 + i) is not a root given equation.