answer:Let S be the value of the given expression. Using sum and difference of cubes to factor, we get [begin{aligned} S &= dfrac{(2-1)(2^2+2+1)}{(2+1)(2^2-2+1)}cdotdfrac{(3-1)(3^2+3+1)}{(3+1)(3^2-3+1)} cdotdfrac{(4-1)(4^2+4+1)}{(4+1)(4^2-4+1)}cdotdfrac{(5-1)(5^2+5+1)}{(5+1)(5^2-5+1)}cdotdfrac{(6-1)(6^2+6+1)}{(6+1)(6^2-6+1)} &= frac{1}{3} cdot frac{2}{4} cdot frac{3}{5} cdot frac{4}{6} cdot frac{5}{7} cdot frac{2^2+2+1}{2^2-2+1} cdot frac{3^2+3+1}{3^2-3+1} cdot frac{4^2+4+1}{4^2-4+1} cdot frac{5^2+5+1}{5^2-5+1} cdot frac{6^2+6+1}{6^2-6+1}.end{aligned}]The first product telescopes to tfrac{1 cdot 2}{6 cdot 7} = tfrac{1}{21}. The second product also telescopes due to the identity [x^2 + x + 1 = (x+1)^2 - (x+1) + 1.]That is, the terms 2^2+2+1 and 3^2-3+1 cancel, as do the terms 3^2+3+1 and 4^2-4+1, and so on, leaving just tfrac{6^2+6+1}{2^2-2+1} = tfrac{43}{3}. Thus, [S = frac{1}{21} cdot frac{43}{3} = frac{43}{63}.]

answer:Let K be the area of the triangle, and let p be the semi-perimeter. Then by Heron's formula, [K^2 = p(p - r)(p - s)(p - t).]By Vieta's formulas, r + s + t = 4, so p = 2. Also, since r, s, t are the roots of x^3 - 4x^2 + 5x - frac{19}{10}, [x^3 - 4x^2 + 5x - frac{19}{10} = (x - r)(x - s)(x - t).]Setting x = 2, we get [(2 - r)(2 - s)(2 - t) = frac{1}{10}.]Then [K^2 = 2(2 - r)(2 - s)(2 - t) = frac{1}{5},]so K = sqrt{frac{1}{5}} = frac{sqrt{5}}{5}.

answer:Recall that a parabola is defined as the set of all points that are equidistant to the focus F and the directrix. To make the algebra a bit easier, we can find the directrix of the parabola y = 8x^2, and then shift it upward 2 units to find the directrix of the parabola y = 8x^2 + 2. Since the parabola y = 8x^2 is symmetric about the y-axis, the focus is at a point of the form (0,f). Let y = d be the equation of the directrix. [asy] unitsize(1.5 cm); pair F, P, Q; F = (0,1/4); P = (1,1); Q = (1,-1/4); real parab (real x) { return(x^2); } draw(graph(parab,-1.5,1.5),red); draw((-1.5,-1/4)--(1.5,-1/4),dashed); draw(P--F); draw(P--Q); dot("F", F, NW); dot("P", P, E); dot("Q", Q, S); [/asy] Let (x,8x^2) be a point on the parabola y = 8x^2. Then [PF^2 = x^2 + (8x^2 - f)^2]and PQ^2 = (8x^2 - d)^2. Thus, [x^2 + (8x^2 - f)^2 = (8x^2 - d)^2.]Expanding, we get [x^2 + 64x^4 - 16fx^2 + f^2 = 64x^4 - 16dx^2 + d^2.]Matching coefficients, we get begin{align*} 1 - 16f &= -16d, f^2 &= d^2. end{align*}From the first equation, f - d = frac{1}{16}. Since f^2 = d^2, f = d or f = -d. We cannot have f = d, so f = -d. Then -2d = frac{1}{16}, so d = -frac{1}{32}. Thus, the equation of the directrix of y = 8x^2 is y = -frac{1}{32}, so the equation of the directrix of y = 8x^2 + 2 is y = frac{63}{32}.

answer:Let alpha = a + bi and gamma = c + di, where a, b, c, and d are real numbers. Then begin{align*} f(1) &= (4 + i) + alpha + gamma = (a + c + 4) + (b + d + 1)i, f(i) &= (4 + i)(-1) + alpha i + gamma = (-b + c - 4) + (a + d - 1)i. end{align*}Since f(1) and f(i) are both real, b + d + 1 = 0 and a + d - 1 = 0, so a = -d + 1 and b = -d - 1. Then begin{align*} |alpha| + |gamma| &= sqrt{a^2 + b^2} + sqrt{c^2 + d^2} &= sqrt{(-d + 1)^2 + (-d - 1)^2} + sqrt{c^2 + d^2} &= sqrt{2d^2 + 2} + sqrt{c^2 + d^2} &ge sqrt{2}. end{align*}Equality occurs when a = 1, b = -1, c = 0, and d = 0. Therefore, the minimum value is sqrt{2}.