\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
We have \begin {align*} a(x - \alpha )^3 + b(x - \beta )^3 & = ax^3 - 3a\alpha x^2 + 3a\alpha ^2 x - a\alpha ^3 + bx^3 - 3b\beta x^2 + 3b\beta ^2 x - b\beta ^3 \\ & = (a + b)x^3 - 3(a\alpha + b\beta ) x^2 + 3 (a\alpha ^2 + b\beta ^2) x - (a\alpha ^3 + b\beta ^3). \end {align*}
By comparing coefficients, we have \[ \left \{ \begin {aligned} a + b & = 1, \\ -3 (a\alpha + b\beta ) = 0 \implies a\alpha + b\beta & = 0, \\ 3 (a\alpha ^2 + b\beta ^2) = -3p \implies a\alpha ^2 + b\beta ^2 & = -p, \\ -(a\alpha ^3 + b\beta ^3) = q \implies a\alpha ^3 + b\beta ^3 & = -q. \end {aligned} \right . \]
The first pair of equation solve to \[ (a, b) = \left (-\frac {\beta }{\alpha - \beta }, \frac {\alpha }{\alpha - \beta }\right ). \]
Putting this into the third equation, we can see \begin {align*} \LHS & = \frac {\beta }{\beta - \alpha } \cdot \alpha ^2 - \frac {\alpha }{\beta - \alpha } \cdot \beta ^2 \\ & = \frac {\alpha \beta (\alpha - \beta )}{\beta - \alpha } \\ & = -\alpha \beta \\ & = - \frac {p^2}{p} \\ & = -p \\ & = \RHS , \end {align*}
using Vieta’s Theorem for \(\alpha \beta \), and for the final one, \begin {align*} \LHS & = \frac {\beta }{\beta - \alpha } \cdot \alpha ^3 - \frac {\alpha }{\beta - \alpha } \cdot \beta ^3 \\ & = \frac {\alpha \beta (\alpha ^2 - \beta ^2)}{\beta - \alpha } \\ & = - \frac {\alpha \beta (\alpha + \beta ) (\beta - \alpha )}{\beta - \alpha } \\ & = -\alpha \beta (\alpha + \beta ) \\ & = -\frac {p^2}{p} \cdot \left (- \frac {-q}{p}\right ) \\ & = -p \cdot \frac {q}{p} \\ & = -q \\ & = \RHS , \end {align*}
using Vieta’s Theorem for \(\alpha \beta \) and \(\alpha + \beta \). Hence, this means for \(\alpha , \beta \) being solutions to \(pt^2 - qt + p^2 = 0\) and \[ (a, b) = \left (-\frac {\beta }{\alpha - \beta }, \frac {\alpha }{\alpha - \beta }\right ), \] we have \[ x^3 - 3px + q = a (x - \alpha )^3 + b (x - \beta )^3. \]
In this case here, we have \(p = 8\) and \(q = 48\). Hence, the quadratic equation is \[ 8t^2 - 48t + 8^2 = 8 (t^2 - 6t + 8) = 8 (t - 2)(t - 4) = 0, \] which solves to \((\alpha , \beta ) = (2, 4)\) or \((\alpha , \beta ) = (4, 2)\). Without loss of generality, let \((\alpha , \beta ) = (2, 4)\), and hence \[ (a, b) = \left (-\frac {\beta }{\alpha - \beta }, \frac {\alpha }{\alpha - \beta }\right ) = \left (-\frac {4}{2 - 4}, \frac {2}{2 - 4}\right ) = \left (2, -1\right ), \]
Hence, the original cubic equation \[ x^3 - 24x + 48 = 0 \] can be simplified to \[ 2 (x - 2) ^ 3 - (x - 4)^3 = 0. \]
Hence, \[ 2(x - 2)^3 = (x - 4)^3, \] and we have \[ 2^{\frac {1}{3}} (x - 2) = \omega ^n (x - 4), \] for \(n = 0, 1, 2\) and \(\omega = \exp \left (\frac {2\pi i}{3}\right )\).
Rearranging gives us \[ x = \frac {2 \left (2\omega ^n - 2^{\frac {1}{3}}\right )}{\omega ^n - 2^{\frac {1}{3}}} \]
When \(n = 0\), \(\omega ^n = 1\), and hence \[ x = \frac {2 \left (2 - 2^{\frac {1}{3}}\right )}{1 - 2^{\frac {1}{3}}}. \]
The other two solutions \[ x = \frac {2 \left (2 \omega - 2^{\frac {1}{3}}\right )}{\omega - 2^{\frac {1}{3}}}, x = \frac {2 \left (2 \omega ^2 - 2^{\frac {1}{3}}\right )}{\omega ^2 - 2^{\frac {1}{3}}}. \]
This equation reduces to \[ x^3 - 3r^2 x + 2 r^3 = 0. \]
This can be factorised to \[ (x - r)(x^2 + rx - 2r^2) = (x - r)^2 (x + 2r) \] and the solutions are \[ x_{1, 2} = r, x_3 = -2r. \]