If $y=1+\sqrt{3}+\sqrt{4}$, then the value of $2 y^4-8 y^3-6 y^2+28 y-84$ is:

$40\sqrt{3}$

(x + y)² = x² + y² +2xy

= y = 1 + √3 + √4,

= y = 1 + √3 + 2 = 3 + √3

= y² = (3 + √3)²

= y² = 9 +3 + 6√3 = 12 + 6√3

According to the question,

= 2y⁴ - 8y³ - 6y² + 28y - 84

= 2y²(y² - 4y - 3) + 28(3 + √3) - 84

= 2y²(y² - 4y - 3 + 4 - 4) + 84 + 28√3 - 84

= 2y²(y² - 4y + 4 - 7) + 28√3

= 2y²{(y-2)² - 7} + 28√3

Now put the value of y and y²

= 2(12 + 6√3){(3 + √3 -2)² - 7} + 28√3

= (24 + 12√3){(1 + √3)2 - 7} + 28√3

= (24 + 12√3){1 + 3 + 2√3 - 7} + 28√3

= (24 + 12√3){2√3 - 3} + 28√3

= 48√3 - 72 + 72 - 36√3 + 28√3

= 40√3

Option 1 is right.