Question

разложите на множители
A) 1-125C^3=
Б)t^3+27=
B)-64q^12+27q^9
Г)(x-1)^3+8
Д)125с^3+27
Е)64-b^3
Ж)q^9-8q^3
З)1+(y+3)^3

Answer 1

$1-125c^3=1-(5c)^3=\left(1-5c\right)\left(1+5c+25c^2\right)\\\\t^3+27=t^3+3^3=\left(t+3\right)\left(t^2-3t+9\right)\\\\-64q^{12}+27q^9=-q^9\left(64q^3-27\right)=-q^9\left(4q-3\right)\left(16q^2+12q+9\right)\\\\(x-1)^3+8=(x-1)^3+2^3=(x-1+2)\left((x-1)^2-2(x-1)+4\right)=(x+1)(x^2-2x+1-2x+2+4)=(x+1)(x^2-4x+7)$

$125c^3+27=\left(5c\right)^3+3^3=\left(5c+3\right)\left(25c^2-15c+9\right)\\\\64-b^3=4^3-b^3=(4-b)(16+4b+b^2)\\\\q^9-8q^3=q^3\left(q^6-8\right)=q^3\left((q^2)^3-2^3\right)=q^3\left(q^2-2\right)\left(q^4+2q^2+4\right)\\\\1+\left(y+3\right)^3=1^3+\left(y+3\right)^3=(1+y+3)\left(1-(y+3)+(y+3)^2\right)=(y+4)(1-y-3+y^2+6y+9)=(y+4)(y^2+5y+7)$