Question

Разложите на множители.

25a³-1
125+b³
1-36y³
c³-4ac+4a³

Answer 1

25a³-1=$( \sqrt[3]{25} a-1)*( \sqrt[3]{25^2} a^2+ \sqrt[3]{25} a+1)$
125+b³=(5+b)(25-5b+b²
1-36y³=$(1- \sqrt[3]{36} y)(1+ \sqrt[3]{36} y+36 \sqrt[3]{36} y^3)$

Answer 2

25a³-1 =(∛25a)³ -1³= (∛25a-1)((∛25a)²-∛25a+1)

125+b³=5³+b³ =(5+b)(25-5b+b²)

1-36y³=1³-(∛36y)³= (1-∛36y)(1+∛36y+(∛36y)²)

c³-4ac+4a³=c³-2ac-2ac+4a³=c(c²-2a)-2a(c-2a²)