Question

Разложение на множители суммы и разности кубов
m³n³-1=
1+q³p³=
27-a³b³=
p⁶-q³=
m³+n¹²=
x³-y⁶=

Answer 1

m³n³-1=(mn)³-1³=(mn-1)((mn)²+mn×1+1²)=(mn-1)(m²n²+mn+1)

1+q³p³=(1+qp)(1-pq+q²p²)=(1+pq)(1-pq+p²q²)

27-a³b³=3³-(ab)³=(3-ab)(3²+3ab+(ab)²)=(3-ab)(9+3ab+a²b²)

p⁶-q³=p³ˣ²-q³=(p²)³-q³=(p²-q)((p²)²+p²q+q²)=(p²-q)(p⁴+p²q+q²)

m³+n¹²=m³+n³ˣ⁴=m³+(n⁴)³=(m+n⁴)(m²-mn⁴+(n⁴)²)=(m+n⁴)(m²-mn⁴+n⁸)

x³-y⁶=x³-y³ˣ²=x³-(y²)³=(x-y²)(x²+xy²+(y²)²)=(x-y²)(x²+xy²+y⁴).