Question

Show that x^3 - 6x - 6 = 0 if x = ∛2 + ∛4

Answer 1

Answer: x³ - 6x - 6 = 0

Step-by-step explanation:

x = ∛2 + ∛4

x = ∛2 + (∛2)²

Let y = ∛2,

=> x = y + y²

raising both sides to the power of 3,

=> x³ = (y² + y)³

x³ = y^6 + 3y^5 + 3y^4 + y³

Substituting y with ∛2

x³ = (∛2)^6 + 3(∛2)^5 + 3(∛2)^4 + (∛2)³

x³ = 2² + 3(∛2)^5 + 3(∛2)^4 + 2¹

x³ = 6 + 3[(∛2)^5 + 3(∛2)^4]

x³ = 6 + 3[2{(∛2)² + ∛2}]

x³ = 6 + 6{(∛2)² + ∛2}

But (∛2)² + ∛2 = x, hence

x³ = 6 + 6x, and

x³ - 6x - 6 = 0