Penjelasan: Awali dengan permisalan [tex]a=3+2\sqrt{2},\:\:\:b=3-2\sqrt{2}[/tex] karena kesekawanan [tex]\displaystyle b=\frac{1}{a}[/tex] begitupun sebaliknya untuk a, maka [tex]\displaystyle a^{-3}+b^{-3}=\frac{1}{a^3}+\frac{1}{b^3}=\frac{a^3+b^3}{(ab)^3}\\\\=\frac{(a+b)(a^2+b^2-ab)}{(ab)^3}\\\\=\frac{(a+b)((a+b)^2-2ab-ab)}{(ab)^3}\\\\=\frac{(a+b)((a+b)^2-3ab)}{(ab)^3}\\\\=\frac{\left(a+\frac{1}{a}\right)\left(\left(a+\frac{1}{a}\right)^2-3a\left(\frac{1}{a}\right)\right)}{\left(\not a\left(\frac{1}{\not a}\right)\right)^3}\\\\=\frac{\left(a+\frac{1}{a}\right)\left(\left(a+\frac{1}{a}\right)^2-3\right)}{1}[/tex]
Verified answer
Jawab:
= 198
Penjelasan:
Awali dengan permisalan
[tex]a=3+2\sqrt{2},\:\:\:b=3-2\sqrt{2}[/tex]
karena kesekawanan
[tex]\displaystyle b=\frac{1}{a}[/tex]
begitupun sebaliknya untuk a,
maka
[tex]\displaystyle a^{-3}+b^{-3}=\frac{1}{a^3}+\frac{1}{b^3}=\frac{a^3+b^3}{(ab)^3}\\\\=\frac{(a+b)(a^2+b^2-ab)}{(ab)^3}\\\\=\frac{(a+b)((a+b)^2-2ab-ab)}{(ab)^3}\\\\=\frac{(a+b)((a+b)^2-3ab)}{(ab)^3}\\\\=\frac{\left(a+\frac{1}{a}\right)\left(\left(a+\frac{1}{a}\right)^2-3a\left(\frac{1}{a}\right)\right)}{\left(\not a\left(\frac{1}{\not a}\right)\right)^3}\\\\=\frac{\left(a+\frac{1}{a}\right)\left(\left(a+\frac{1}{a}\right)^2-3\right)}{1}[/tex]
[tex]\displaystyle=\left(a+\frac{1}{a}\right)\left(\left(a+\frac{1}{a}\right)^2-3\right)\\\\=\left(a+b\right)\left(\left(a+b\right)^2-3\right)[/tex]
Cari a + b
[tex]\displaystyle a+b=3+2\sqrt{2}+3-2\sqrt{2}\\a+b=6[/tex]
Oleh karena itu
[tex]\left(a+b\right)\left(\left(a+b\right)^2-3\right)[/tex]
= 6(6² - 3)
= 6(36 - 3)
= 6(33)
= 198
(xcvi)