The equation $4k^3 - 27k + k = 0$ has a double root - HSC - SSCE Mathematics Extension 2 - Question 5 - 2002 - Paper 1

Taking the given roots $\alpha$, $\beta$, and $\gamma$ of the polynomial $x^3 - 5x^2 + 5 = 0$, we substitute $x = y + 1$:

$(y + 1)^3 - 5(y + 1)^2 + 5 = 0$ Expanding this expression results in:

$y^3 + 3y^2 + 3y + 1 - 5(y^2 + 2y + 1) + 5 = 0$ This simplifies to: $y^3 - 2y^2 - 2 = 0$ Thus the polynomial we seek is: $y^3 - 2y^2 - 2 = 0$

Using $\alpha^2$, $\beta^2$, and $\gamma^2$, we apply the transformation method. From the equation $x^3 - 5x^2 + 5 = 0$, we can find the sum of squares:

Using the identity for the roots:

$\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)$ Calculating gives: $\alpha^2 + \beta^2 + \gamma^2 = 25 - 2(5) = 15$ We then set up the polynomial in $x$: $x^3 - 15x + \text{ constant} = 0$ The actual constant can be derived using the product of roots, arriving at: $x^3 - 15x + 25 = 0$.

To find $\alpha^3 + \beta^3 + \gamma^3$, we can use the relationship:

$\alpha^3 + \beta^3 + \gamma^3 = (\alpha + \beta + \gamma)(\alpha^2 + \beta^2 + \gamma^2 - \alpha\beta - \beta\gamma - \gamma\alpha) + 3\alpha\beta\gamma$ Substituting the known values: $\alpha + \beta + \gamma = 5, \quad \alpha^2 + \beta^2 + \gamma^2 = 15, \quad \text{and} \quad \alpha\beta\gamma = -5$ This gives: $\alpha^3 + \beta^3 + \gamma^3 = 5(15 - 5) + 3(-5) = 50 - 15 = 35$.

The general formula for the tangent line to an ellipse at the point $(x_p, y_p)$ is given by:

$\frac{x}{a^2} y_p + \frac{y}{b^2} x_p = 1$ Substituting point $(x_p, y_p)$ into the standard ellipse equation confirms that the point lies on the ellipse and thus the tangent line is formulated correctly.

Using the concept of the chord of contact from point $T(x_0, y_0)$ to the ellipse, the equation can be expressed as:

$\frac{x_0}{a^2} + \frac{y_0}{b^2} = 1$ This is derived by considering the slopes from point T to various points on the ellipse and ensuring the conditions are satisfied, confirming that T lies outside the ellipse.