<s>[INST] Lagrange multipliers to optimize sum of cubes of roots of a quadratic 
If $\lambda \geq -2$ such that the sum of cubes of roots of the equation $x^2 + (2-\lambda )x + (10-\lambda) = 0$ is minimum, then the magnitude of difference of root is..?

Let roots be $\alpha $ and $\beta$, we wish to maximize $f(\alpha,\beta) = \alpha^3 + \beta^3$ under the constraint of $ \alpha + \beta = 2- \lambda $ and $\alpha \beta= 10 -\lambda$ , combining the constraints we get:
$$ \alpha + \beta + \alpha \beta = 8$$
Now apply lagrange multiplier equation, we get:
$$ 3 \alpha^2 = \mu ( 1 + \beta)$$
And,
$$ 3 \beta^2 = \mu ( 1 + \alpha)$$
Taking the ratio of the two equations:
$$ \frac{\alpha^2 }{\beta^2} = \frac{1+\beta}{1 + \alpha} \tag{1}$$
Substracting one on both sides:
$$ \frac{\alpha + \beta}{\beta^2} =  \frac{-1}{1+\alpha} \tag{2}$$
This leads to:
$$ \alpha + \beta + \alpha^2 + \alpha \beta  = - \beta^2$$
Hence,
$$ -12 = \alpha^2 + \beta^2$$
Now.. I am not sure what to do. One way is to ditch the lagrange multiplier approach and directly express the objective function in terms of $\lambda$ and minimize but is there a way to proceed further from the information we have?
 [/INST] This is a solution without Lagrange Multiplier method. See my edit in the end for Lagrange Multiplier method.
$\alpha + \beta = \lambda-2, \alpha\beta = 10 - \lambda$
$f(\alpha, \beta) = \alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3 \alpha \beta(\alpha + \beta)$
So, $f(\alpha, \beta) = (\lambda-2)^3 - 3 (10-\lambda)(\lambda-2) = \lambda^3 - 3 \lambda^2 - 24 \lambda + 52$
Now we take derivative with respect to $\lambda$ and equate to $0$,
$\lambda^2 - 2\lambda-8 = 0 \implies \lambda = -2, 4$
Second derivative test confirms a local minima at $\lambda = 4$. We also observe that the function $f$ is monotonically decreasing for $\lambda \lt - 2$ and is monotonically increasing for $\lambda \gt 4$. So in the given domain $\lambda \geq - 2$, absolute minima occurs at $\lambda = 4$.
And so we have,
$\alpha + \beta = \lambda - 2 = 2, \alpha \beta = 10 - \lambda = 6$
Solving, $\alpha = 1 \pm i \sqrt 5, \beta = 1 \mp i \sqrt5$, which are complex roots of the quadratic for $\lambda = 4$.

For Lagrange Multiplier method,
Say $\alpha + \beta = x, \alpha \beta = y$
You found $\alpha^2 + \beta^2 = - 8$. We also have $y = 8 - x$
$x^2 = - 8 + 2y = - 8 + 16 - 2x \implies x^2 + 2x - 8 = 0 \ $
(where $x = \alpha + \beta$)
Solving $x = 2, y = 6; x = - 4, y = 12$.
We test $\alpha^3 + \beta^3 = x^3 - 3 xy$ for both and obtain $x = 2, y = 6$ to be the minima.
$(\alpha-\beta)^2 = x^2 - 4y = - 20 \implies \alpha-\beta = \pm \sqrt{-20}$
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