\(% 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}}\)
Notice that \[ (1+ax)(1+bx)(1+cx) = 1 + (a + b + c) x + (ab + ac + bc) x^2 + abc x^3, \] and by comparing coefficients we have \[ q = bc + ca + ab, r = abc \]
Using the identities for the logarithms, we have \begin {align*} \ln (1 + qx^2 + rx^3) & = \ln (1 + ax) + \ln (1 + bx) + \ln (1 + cx) \\ & = \sum _{k = 1}^{\infty } (-1)^{k + 1} \frac {(ax)^k}{k} + \sum _{k = 1}^{\infty } (-1)^{k + 1} \frac {(bx)^k}{k} + \sum _{k = 1}^{\infty } (-1)^{k + 1} \frac {(cx)^k}{k} \\ & = \sum _{k = 1}^{\infty } (-1)^{k + 1} x^k \frac {a^k + b^k + c^k}{k}, \end {align*}
and hence \[ S_k = \frac {a^k + b^k + c^k}{k}, \] as desired.
Since \begin {align*} S_2 & = \frac {a^2 + b^2 + c^2}{2} \\ & = \frac {(a + b + c)^2 - 2(ab + bc + ca)}{2} \\ & = \frac {0^2 - 2q}{2} \\ & = -q, \end {align*}
\begin {align*} S_3 & = \frac {a^3 + b^3 + c^3}{3} \\ & = \frac {(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) + 3abc}{3} \\ & = abc \\ & = r, \end {align*}
and \begin {align*} S_5 & = \frac {a^5 + b^5 + c^5}{5} \\ & = \frac {(a^2 + b^2 + c^2)(a^3 + b^3 + c^3) - a^2b^2 (a + b) - a^2c^2 (a + c) - b^2 c^2 (b + c)}{5} \\ & = \frac {(-2q)(3r) + a^2 b^2 c + b^2 c^2 a + a^2 c^2 b}{5} \\ & = \frac {-6qr + abc(ab + bc + ac)}{5} \\ & = \frac {-6qr + qr}{5} \\ & = -qr. \end {align*}
Therefore, \(S_2 S_3 = S_5\) as desired.
Notice that \begin {align*} S_7 & = \frac {a^7 + b^7 + c^7}{7} \\ & = \frac {(a^2 + b^2 + c^2)(a^5 + b^5 + c^5)}{7} \\ & = \frac {(-2q) \cdot (-5qr) - a^2 b^2 (a^3 + b^3) - b^2 c^2 (b^3 + c^3) - a^2 c^2 (a^3 + c^3)}{7} \\ & = \frac {10q^2 r - a^2 b^2 (3r - c^3) - b^2 c^2 (3r - a^3) - a^2 c^2 (3r - b^3)}{7} \\ & = \frac {10q^2 r - 3r (a^2b^2 + b^2 c^2 + a^2 c^2) + a^2 b^2 c^2 (a + b + c)}{7} \\ & = \frac {10q^2 r - 3r\left [(ab + bc + ac)^2 - 2abc(a + b + c)\right ] + r^2 \cdot 0}{7} \\ & = \frac {10q^2 r - 3q^2r}{7} \\ & = q^2r. \end {align*}
Also, \(S_2 S_5 = (-q) \cdot (-qr) = q^2 r\), so \(S_2 S_5 = S_7\) as desired.