| b3 | a3 | 1 | a+b | |||||
mam takie cos ( | − | ) * | − ( | )2 | ||||
| 3 | 3 | b−a | 2 |
| (b−a)2 | ||
Jak to rozpisać żeby wynik wyszedl | ||
| 12 |
| (b−a)(a2+ab+b2) | 1 | ||
* | −...= i dalej próbuj | ||
| 3 | b−a |
| b3 − a3 | 1 | a2 + 2ab + b2 | |||
* | − | = | |||
| 3 | b − a | 4 |
| (b − a)(b2 + ab + a2 | 1 | a2 + 2ab + b2 | ||||
= | * | − | = | |||
| 3 | b − a | 4 |
| b2 + ab + a2 | ||
= | − U[a2 + 2ab + b2}{4} = | |
| 3 |
| 4b2 + 4ab + 4a2 | 3a2 + 6ab + 3b2 | |||
= | − | = | ||
| 12 | 12 |
| b2 − 2ab + a2 | (b − a)2 | |||
= | = | |||
| 12 | 12 |
| (b−a)(b2 + ab + a2) | 1 | a2 +2ab +b2 | ||||
= | * | − | = | |||
| 3 | b−a | 4 |
| b2 + ab + a2 | a2 +2ab +b2 | |||
− | = | |||
| 3 | 4 |
| 4(b2 + ab + a2) | 3(a2 +2ab +b2) | ||
− | = | ||
| 12 | 12 |
| b2 − 2ab +a2 | (b−a)2 | ||
= | |||
| 12 | 12 |