Question

помогите пожалуйста мучаюсь уже полтора часа​

Answer 1

Ответ:

Объяснение:

воспользуемся формулами

x³ - y³ = (x - y)(x² + xy + y²)

x³ + y³ = (x + y)(x² - xy + y²)

$\displaystyle\\a^3-27b^3=a^3-(3b)^3=(a-3b)(a^2+3ab+9b^2)\\\\\\k^6+(pq)^6=(k^2)^3+((pq)^2)^3=(k^2+(pq)^2)(k^4-k^2\cdot(pq)^2+(pq)^4)=\\\\\\=(k^2+p^2q^2)(k^4-k^2p^2q^2+p^4q^4)$

Answer 2

Ответ:

$\\ a {}^{3} - 27b {}^{3} = a {}^{3} - 3 {}^{3} b {}^{3} = a {}^{3} - (3b) {}^{3} = (a - 3b) \times (a {}^{2} + a \times 3b + (3b) {}^{2} ) = (a - 3b) \times (a {}^{2} + 3ab + 9b {}^{2} )$

$\\ k {}^{6} + (pq) {}^{6} = (k {}^{2}) {}^{3} + ((pq) {}^{2}) {}^{3} \\ \\ a {}^{3} + b {}^{3} = (a + b)(a {}^{2} - ab + b {}^{2} ) \\ \\ (k {}^{2} + (pq) {}^{2} ) \times (k {}^{4} - k {}^{2} \times (pq) {}^{2} + (pq) {}^{4} ) = (k {}^{2} + p {}^{2} q {}^{2} ) \times (k {}^{4} - k {}^{2} p {}^{2} q {}^{2} + p {}^{4} q {}^{4} )$