Jawab:= 198
Penjelasan :Permisalana = √2 + 1, b = √2 - 1
Karena kesekawananb = 1/a, a = 1/b
a⁶ + b⁶= (a²)³ + (b²)³
Permisalan lagic = a², d = b²
= c³ + d³= (c + d)(c² + d² - cd)= (c + d)((c+d)² - 2cd - cd)= (c + d)((c+d)² - 3cd)
Substitute back= (a²+b²)((a²+b²)² - 3a²b²)= (a²+b²)((a²+b²)² - 3(ab)²)
Karena b = 1/a, dan a = 1/b= (a²+(1/a)²)((a²+(1/a)²)² - 3(a(1/a))²)= (a²+(1/a)²)((a²+(1/a)²)² - 3)Masukkan lagi= (a²+b²)((a²+b²)² - 3)
Lalu cari nilai a² + b²a² + b² = (√2 + 1)² + (√2 - 1)²a² + b² = 2+2√2+1+2-2√2+1a² + b² = 2+1+2+1a² + b² = 6
Maka(a²+b²)((a²+b²)² - 3)= 6(6² - 3)= 6(36-3)= 6(33)= 198
Lah jadi (√2 + 1)⁶ + (√2 - 1)⁶ = (3 + 2√2)⁻³ + (3 - 2√2)⁻³ ?(xcvi)
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Verified answer
Jawab:
= 198
Penjelasan :
Permisalan
a = √2 + 1, b = √2 - 1
Karena kesekawanan
b = 1/a, a = 1/b
a⁶ + b⁶
= (a²)³ + (b²)³
Permisalan lagi
c = a², d = b²
= c³ + d³
= (c + d)(c² + d² - cd)
= (c + d)((c+d)² - 2cd - cd)
= (c + d)((c+d)² - 3cd)
Substitute back
= (a²+b²)((a²+b²)² - 3a²b²)
= (a²+b²)((a²+b²)² - 3(ab)²)
Karena b = 1/a, dan a = 1/b
= (a²+(1/a)²)((a²+(1/a)²)² - 3(a(1/a))²)
= (a²+(1/a)²)((a²+(1/a)²)² - 3)
Masukkan lagi
= (a²+b²)((a²+b²)² - 3)
Lalu cari nilai a² + b²
a² + b² = (√2 + 1)² + (√2 - 1)²
a² + b² = 2+2√2+1+2-2√2+1
a² + b² = 2+1+2+1
a² + b² = 6
Maka
(a²+b²)((a²+b²)² - 3)
= 6(6² - 3)
= 6(36-3)
= 6(33)
= 198
Lah jadi (√2 + 1)⁶ + (√2 - 1)⁶
= (3 + 2√2)⁻³ + (3 - 2√2)⁻³ ?
(xcvi)