![a_{2}=a_{1}+d](https://tex.z-dn.net/?f=a_%7B2%7D%3Da_%7B1%7D%2Bd+)
![a_{3}=a_{1}+2d](https://tex.z-dn.net/?f=a_%7B3%7D%3Da_%7B1%7D%2B2d+)
Найдем второй член арифметической прогрессии из суммы заданной по условию
![S_{3}= \frac{ a_{1}+a_{3}}{2}*3](https://tex.z-dn.net/?f=+S_%7B3%7D%3D+%5Cfrac%7B+a_%7B1%7D%2Ba_%7B3%7D%7D%7B2%7D%2A3++)
⇒
![9= \frac{ a_{1}+a_{1}+2d}{2}*3](https://tex.z-dn.net/?f=9%3D+%5Cfrac%7B+a_%7B1%7D%2Ba_%7B1%7D%2B2d%7D%7B2%7D%2A3)
⇒
![a_{1}+d=3](https://tex.z-dn.net/?f=a_%7B1%7D%2Bd%3D3)
Соответственно
![a_{1}=3-d](https://tex.z-dn.net/?f=a_%7B1%7D%3D3-d)
и
![a_{3}=3+d](https://tex.z-dn.net/?f=a_%7B3%7D%3D3%2Bd)
По условию
(3-d)²+3²+(3+d)²=99
9-6d+d²+9+9+6d+d²=99
2d²=72 ⇒ d=<span>±6 так как прогрессия убывающая, а второй член прогрессии положительный то d=-6
</span>
![a_{1}=3-(-6)=9](https://tex.z-dn.net/?f=a_%7B1%7D%3D3-%28-6%29%3D9)
![a_{5}=a_{1}+4d=9+4(-6)=9-24=-15](https://tex.z-dn.net/?f=a_%7B5%7D%3Da_%7B1%7D%2B4d%3D9%2B4%28-6%29%3D9-24%3D-15)
А^3-в^3=(а-в)(а^2+ав+в^2)
((5k-5) / (k²-1) - k / (k+1)) : (5-k) / (k+1) = ((5k-5) / (k-1)(k+1) - k / (k+1)) :
(5-k) / (k+1) = ((5k-5) / (k-1)(k+1) - (k²-k) / (k-1)(k+1)) : (5-k) / (k+1) =
(5k-5-k²+k) / (k-1)(k+1) : (5-k) / (k+1) = (6k-k²-5) / (k-1)(k+1) : (5-k) / (k+1) =
(6k-k²-5)(k+1) / (k-1)(k+1)(5-k) = (6k-k²-5) / (k-1)(5-k) = (6k-k²-5) /
(5k-k²-5+k) = (6k-k²-5) / (6k-k²-5) = 1.