(b) $f(x) = 2(x – 3)(x – 5)$; $\left| {\,\begin{array}{*{20}{c}}1&{x + 3}&{3({x^2} + 3x + 9)}\\1&{x + 5}&{4({x^2} + 5x + 25)}\\1&1&3\end{array}\,} \right|$

(Taking out $(x – 3),(x – 5)$ and $2$ from $I^{st}$ row, IInd row and $II^{rd}$ column respectively)

$f(x) = 2(x – 3)(x – 5)$

$\left| {\,\begin{array}{*{20}{c}}0&{(x + 2)}&{3({x^2} + 3x + 8)}\\0&2&{{x^2} + 11x + 73}\\1&1&3\end{array}\,} \right|,\,\,$

$({R_1} \to {R_1} – {R_3}$ and ${R_2} \to {R_2} – {R_1}$)

$ = 2(x – 3)(x – 5)[1(x + 2)$$({x^2} + 11x + 73) – 6({x^2} + 3x + 8)]$

$ = 2({x^2} – 8x + 15)({x^3} + 13{x^2} + 95x$$ + 146 – 6{x^2} – 18x – 48)$

$ = 2({x^2} – 8x + 15)({x^3} + 7{x^2} + 77x + 98)$

$ = 2({x^5} – {x^4} + 36{x^3} – 413{x^2}$$ + 371x + 1470)$

$f(1) = 2928$,$f(3) = 0$,$f(5) = 0$

 $f(1).f(3) + f(3).f(5) + f(5).f(1)$ $ = 0 + 0 + 0$ $ = 0 = f(3)$.