Задание 465

Решение:

• \({\largeа)}\ a^3-27=a^3-3^3=(a-3)(a^2+3\cdot{a}+3^2)=(a-3)(a^2+3a+9)\)

• \({\largeб)}\ 27+8x^3=3^3+(2x)^3=(3+2x)(3^2-3\cdot2x+(2x)^2)=(3+2x)(9-6x+4x^2)\)

• \({\largeв)}\ 8m^3-n^3=(2m)^3-n^3=(2m-n)((2m)^2+2m\cdot{n}+n^2)=(2m-n)(4m^2+2mn+n^2)\)

• \({\largeг)}\ 1+y^6=1^3+(y^2)^3=(1+y^2)(1^2-1\cdot{y}^2+(y^2)^2)=(1+y^2)(1-y^2+y^4)\)

• \({\largeд)}\ x^9-125=(x^3)^3-5^3=(x^3-5)((x^3)^2+x^3\cdot5+5^2)=(x^3-5)(x^6+5x^3+25)\)

• \({\largeе)}\ 64a^3+b^6=(4a)^3+(b^2)^3=(4a+b^2)((4a)^2-4a\cdot{b}^2+(b^2)^2)=(4a+b^2)(16a^2-4ab^2+b^4)\)

• \({\largeж)}\ \frac{1}{8}-m^{12}=\left(\frac{1}{2}\right)^3-(m^4)^3=\left(\frac{1}{2}-m^4\right)\left(\left(\frac{1}{2}\right)^2+\frac{1}{2}\cdot{m}^4+(m^4)^2\right)=\left(\frac{1}{2}-m^4\right)\left(\frac{1}{4}+\frac{1}{2}m^4+m^8\right)\)

• \({\largeз)}\ \frac{8}{27}+n^3=\left(\frac{2}{3}\right)^3+n^3=\left(\frac{2}{3}+n\right)\left(\left(\frac{2}{3}\right)^2-\frac{2}{3}\cdot{n}+n^2\right)=\left(\frac{2}{3}+n\right)\left(\frac{4}{9}-\frac{2}{3}n+n^2\right)\)

• \({\largeи)}\ 0{,}125-27x^3=0{,}5^3-(3x)^3=(0{,}5-3x)((0{,}5)^2+0{,}5\cdot3x+(3x)^2)=(0{,}5-3x)(0{,}25+1{,}5x+9x^2)\)