How to Solve the Equation x^32 x^3 2 0 and Factor it?

To solve the equation x³² – x³ + 2 = 0, we can start by rewriting it in terms of a substitution variable, u. Let’s define u such that:

• u = x³

Now, we can express the equation in terms of u:

• u² – u + 2 = 0

This is a standard quadratic equation. To solve it, we can use the quadratic formula:

• u = (-b ± √(b² – 4ac)) / 2a

In our case:

• a = 1
• b = -1
• c = 2

Now, plugging in these values:

• u = (1 ± √((-1)² – 4 × 1 × 2)) / (2 × 1)
• u = (1 ± √(1 – 8)) / 2
• u = (1 ± √(-7)) / 2

Since the discriminant is negative, we have:

• u = (1 ± i√7) / 2

Thus, we have two complex solutions for u. But remember, we defined u = x³. So, we will now substitute back to find the values of x:

• x³ = (1 + i√7) / 2
• x³ = (1 – i√7) / 2

Now, to find x, we take the cube root of both sides. This will yield:

• x = ∛((1 + i√7) / 2)
• x = ∛((1 – i√7) / 2)

In summary, the equation x³² – x³ + 2 = 0 leads us to two complex solutions for x after substituting and factoring the quadratic. These roots reflect that the original equation does not have real solutions but rather complex ones.