Question

Jika p, q, r akar-akar persamaan x³ - x - 1 = 0, tentukan nilai dari
[tex]\displaystyle \frac{1-p}{1+p}+\frac{1-q}{1+q}+\frac{1-r}{1+r}[/tex]

Answer 1

Verified answer

PERSAMAAN KUBIK

ax³ + bx² + cx + d = 0

dengan akar-akar : p, q, r

p + q + r = -b/a

pq + qr + rp = c/a

pqr = -d/a

•

x³ - x - 1 = 0

x(x² - 1) = 1

x(x - 1) = 1/(x + 1)

•

p + q + r = 0

pq + qr + rp = -1

pqr = 1

p³ + q³ + r³ - 3pqr = (p + q + r)(p² + q² + r² - (pq + qr + rp)) = 0

p³ + q³ + r³ = 3pqr = 3

p² + q² + r²

= (p + q + r)² - 2(pq + qr + rp)

= 0 - 2(-1)

= 2

•

soal

= -p(p - 1)² - q(q² - 1) - r(r - 1)²

= -(p³ + q³ + r³) + 2(p² + q² + r²) - (p + q + r)

= -3 + 2(2) - 0

= 1