\(% 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}}\)

2023.2.4 Question 4

  1. We have \begin {align*} \left (x - \sqrt {2}\right )^2 & = 3 \\ x^2 - 2\sqrt {2}x + 2 & = 3 \\ x^2 - 1 & = 2 \sqrt {2} x \\ x^4 - 2x^2 + 1 & = 8x^2 \\ x^4 - 10 x^2 + 1 & = 0 \end {align*}

    as desired.

    If \(f(x) = x^4 - 10x^2 + 1\), we notice that \(x = \sqrt {2} + \sqrt {3}\) satisfies \(\left (x - \sqrt {2}\right )^2 = \left (\sqrt {3}\right )^2 = 3\), and hence \(f\left (\sqrt {2} + \sqrt {3}\right ) = 0\) as desired.

  2. We have \begin {align*} \left (x - \left (\sqrt {2} + \sqrt {3}\right )\right )^2 & = \left (\sqrt {5}\right )^2 = 5 \\ x^2 - 2 \left (\sqrt {2} + \sqrt {3}\right ) x + 2 + 3 + 2 \sqrt {6} & = 5 \\ x^2 + 2 \sqrt {6} & = 2 \left (\sqrt {2} + \sqrt {3}\right ) x \\ x^4 + 2 \cdot 2 \sqrt {6} \cdot x^2 + \left (2\sqrt {6}\right )^2 & = 4 \left (\sqrt {2} + \sqrt {3}\right )^2 x^2 \\ x^4 + 4 \sqrt {6} x^2 + 24 & = 4 \left (5 + 2\sqrt {6}\right )x^2 \\ x^4 + 4 \sqrt {6} x^2 + 24 & = 20 x^2 + 8 \sqrt {6} x^2 \\ x^4 - 20 x^2 + 24 & = 4 \sqrt {6} x^2 \\ \left (x^4 - 20 x^2 + 24\right )^2 & = \left (4 \sqrt {6} x^2\right )^2 \\ x^8 - 40 x^6 + 448 x^4 - 960 x^2 + 576 & = 96 x^4 \\ x^8 - 40 x^6 + 352 x^4 - 960 x^2 + 576 & = 0. \end {align*}

    Therefore, the polynomial \[ g(x) = x^8 - 40 x^6 + 352 x^4 - 960 x^2 + 576 \] satisfies \(g\left (\sqrt {2} + \sqrt {3} + \sqrt {5}\right ) = 0\) as desired.

  3. If \(t = a, b, c\) are solutions to the cubic equation \(t^3 - 3t + 1 = 0\) in \(t\), then \(t = a + \sqrt {2}, b + \sqrt {2}, c + \sqrt {2}\) are solutions to the cubic equation in \(t\) \begin {align*} \left (y - \sqrt {2}\right )^3 - 3 \left (t - \sqrt {2}\right ) + 1 & = 0 \\ t^3 - 3 \sqrt {2} t^2 + 6t - 2 \sqrt {2} - 3t + 3 \sqrt {2} + 1 & = 0 \\ t^3 + 3t + 1 & = 3 \sqrt {2} t^2 - \sqrt {2} \\ t^6 + 6 t^4 + 2t^3 + 9t^2 + 6t + 1 & = 18 t^4 - 12 t^2 + 2 \\ t^6 - 12 t^4 + 2t^3 + 21 t^2 + 6t - 1 & = 0. \end {align*}

    Therefore, the polynomial \[ h(x) = x^6 - 12 x^4 + 2 x^3 + 21 x^2 + 6x - 1 \] satisfies \(h \left (a + \sqrt {2}\right ) = h \left (b + \sqrt {2}\right ) = h \left (c + \sqrt {2}\right ) = 0\) as desired.

  4. We have \begin {align*} \left (x - \sqrt [3]{2}\right )^3 & = 3 \\ x^3 - 3 \sqrt [3]{2} x^2 + 3 \sqrt [3]{4} x - 2 & = 3 \\ x^3 - 5 & = 3 \sqrt [3]{2} x^2 - 3 \sqrt [3]{4} x \\ x^3 - 5 & = 3 \sqrt [3]{2} x \left (x - \sqrt [3]{2}\right ) \\ x^3 - 5 & = 3 \sqrt [3]{2} x \cdot \sqrt [3]{3} \\ x^3 - 5 & = 3 \sqrt [3]{6} x \\ x^9 - 15 x^6 + 75 x^3 - 125 & = 162 x^3 \\ x^9 - 15 x^6 - 87 x^3 - 125 & = 0. \end {align*}

    Therefore, the polynomial \[ k(x) = x^9 - 15 x^6 - 87 x^3 - 125 = 0 \] satisfies \(k \left (\sqrt [3]{2} + \sqrt [3]{3}\right ) = 0\).