Vieta's Theorem is a good approach!

Relevant wiki: Vieta's Formula Problem Solving - Intermediate

• $a + b + c =$ $\frac{21}{4}$

• $1/a + 1/b + 1/c =$ $\frac{21}{4}$

• $abc = 1$

$1/a + 1/b + 1/c = 21/4$

$(bc+ac+ab)/(abc) = 21/4$ )

$(bc+ac+ab)/1 = 21/4$

$bc + ac + ab = 21/4$

Let a, b, c be the roots of a certain 3-degree polynomial equation. It follows that…

$(x³ - (a+b+c)x² + (ab+ac+bc)x - (abc) = 0$

$x³ - (21/4)x² + (21/4)x - 1 = 0$

$4x³ - 21x² + 21x - 4 = 0$

$(4x³ - 4) + (-21x² + 21x) = 0$

$4(x³ - 1) + (-21x)(x - 1) = 0$

$4(x - 1)(x² + x + 1) - 21x(x - 1) = 0$

$(x - 1)[4(x² + x + 1) - 21x] = 0$

$(x - 1)(4x² + 4x + 4 - 21x) = 0$

$(x - 1)(4x² - 17x + 4) = 0$

$(x - 1)(4x - 1)(x - 4) = 0$

$x = 1, x = 1/4, x = 4.$

Since $a < b < c$ , therefore…

$a = \frac{1}{4}$

$b = 1$

$c = 4$

Just test the first choice and find that the second choice has the same values organized correctly.