soal matematika plz help

Misal : p = √2 - 1 => p^2 = (√2 - 1)^2 = 2 - 2√2 + 1 = 3 - 2√2
a^2x = √2 - 1
a^2x = p ====> a^x = √p
a^3x = a^2x . a^x = p .√p

(a^3x + a^-3x) / (a^x + a^-x)
= (a^3x + 1/(a^3x)) / (a^x + 1/(a^x))
= (p√p + 1/(p√p)) / (√p + 1/√p) . (p√p)/(p√p)
= (p^3 + 1) / (p^2 + p)
= (p + 1)(p^2 - p + 1) / p(p + 1)
= (p^2 - p + 1) / p
= ((3 - 2√2) - (√2 - 1) + 1) / (√2 - 1)
= (5 - 3√2) / (√2 - 1)
= (5 - 3√2) / (√2 - 1) . (√2 + 1)/(√2 + 1)
= (5√2 + 5 - 6 - 3√2) / (2 - 1)
= (2√2 - 1) / 1
= 2√2 - 1

2) x + 1/x = 1
Misal x = a dan 1/x = b
a + b = x + 1/x = 1
ab = x . 1/x = 1

a^2 + b^2 = (a + b)^2 - 2ab = (1)^2 - 2(1) = 1 - 2 = -1
a^3 + b^3 = (a + b)^3 - 3ab(a + b) = 1^3 - 3(1)(1) = 1 - 3 = -2

(a^2 + b^2)(a^3 + b^3) = (-1)(-2)
a^5 + a^2.b^3 + a^3.b^2 + b^5 = 2
a^5 + a^2.b^2 (b + a) + b^5 = 2
a^5 + (ab)^2 (a + b) + b^5 = 2
a^5 + (1)^2 (1) + b^5 = 2
a^5 + 1 + b^5 = 2
a^5 + b^5 = 2 - 1
x^5 + 1/x^5 = 1