PEMBAHASAN
a³ - 1 = (a - 1)(a² + a + 1)
a³ + 1 = (a + 1)(a² - a + 1)
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Misal :
2018 = a
[tex] \displaystyle \frac{2019^2(2018^2-2017)}{(2018^2-1)(2018^3+1)}\times\frac{2017^2(2018^2+2019)}{1-2018^3} \\ \\ \displaystyle = \frac{ {(a + 1)}^{2} ( {a}^{2} - a + 1)}{(a + 1)(a - 1)(a + 1)( {a}^{2} - a + 1 )} \: \times \: \frac{ { (a - 1)^{ 2} ( {a}}^{2} + a + 1 )}{ - ( {a}^{3} - 1)} \\ \\ \displaystyle = \frac{1}{a - 1} \times - (a - 1) \\ \\ = \boxed{ - 1}[/tex]
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Verified answer
PEMBAHASAN
a³ - 1 = (a - 1)(a² + a + 1)
a³ + 1 = (a + 1)(a² - a + 1)
__
Misal :
2018 = a
__
[tex] \displaystyle \frac{2019^2(2018^2-2017)}{(2018^2-1)(2018^3+1)}\times\frac{2017^2(2018^2+2019)}{1-2018^3} \\ \\ \displaystyle = \frac{ {(a + 1)}^{2} ( {a}^{2} - a + 1)}{(a + 1)(a - 1)(a + 1)( {a}^{2} - a + 1 )} \: \times \: \frac{ { (a - 1)^{ 2} ( {a}}^{2} + a + 1 )}{ - ( {a}^{3} - 1)} \\ \\ \displaystyle = \frac{1}{a - 1} \times - (a - 1) \\ \\ = \boxed{ - 1}[/tex]