If a tangent drawn at `P(alpha, alpha^(3))` to the curve `y=x^(3)` meets it again at `Q(beta, beta^(3))`, then `2beta+alpha` is equal to

Options

A. 0
B. `-3alpha`
C. `3alpha`
D. `4alpha`

Answer

To solve the problem step by step, we will follow the reasoning outlined in the video transcript while providing a clear mathematical representation.

### Step 1: Identify the curve and point of tangency
The curve given is \( y = x^3 \). We have a point \( P(\alpha, \alpha^3) \) on this curve.

### Step 2: Find the slope of the tangent at point \( P \)
To find the slope of the tangent at point \( P \), we differentiate the curve:
\[ \frac{dy}{dx} = 3x^2 \]
Now, substituting \( x = \alpha \):
\[ \text{slope at } P = 3\alpha^2 \]

### Step 3: Write the equation of the tangent line at point \( P \)
Using the point-slope form of the equation of a line, the equation of the tangent at point \( P \) is:
\[ y - \alpha^3 = 3\alpha^2(x - \alpha) \]
Rearranging this gives:
\[ y = 3\alpha^2x - 3\alpha^3 + \alpha^3 \]
\[ y = 3\alpha^2x - 2\alpha^3 \]

### Step 4: Find the intersection of the tangent with the curve again
We need to find where this tangent line intersects the curve \( y = x^3 \) again. Set the two equations equal:
\[ 3\alpha^2x - 2\alpha^3 = x^3 \]
Rearranging gives:
\[ x^3 - 3\alpha^2x + 2\alpha^3 = 0 \]

### Step 5: Factor the cubic equation
Since \( x = \alpha \) is a root (because it corresponds to point \( P \)), we can factor the cubic polynomial:
\[ x^3 - 3\alpha^2x + 2\alpha^3 = (x - \alpha)(x^2 + ax + b) \]
Using polynomial long division or synthetic division, we find:
\[ x^2 + \alpha x - 2\alpha^2 \]

### Step 6: Find the other root
To find the roots of the quadratic \( x^2 + \alpha x - 2\alpha^2 = 0 \), we use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-\alpha \pm \sqrt{\alpha^2 + 8\alpha^2}}{2} = \frac{-\alpha \pm 3\alpha}{2} \]
This gives us:
\[ x = \alpha \quad \text{or} \quad x = -2\alpha \]
Since \( x = \alpha \) corresponds to point \( P \), the other intersection point \( Q \) is at:
\[ \beta = -2\alpha \]

### Step 7: Calculate \( 2\beta + \alpha \)
Now, substituting \( \beta = -2\alpha \) into \( 2\beta + \alpha \):
\[ 2\beta + \alpha = 2(-2\alpha) + \alpha = -4\alpha + \alpha = -3\alpha \]

### Final Answer
Thus, the value of \( 2\beta + \alpha \) is:
\[ \boxed{-3\alpha} \]