Задание 408

Решение:

• \({\largeа)}\ (3p-10q)(100q^2+30pq+9p^2)=(3p-10q)(9p^2+30pq+100q^2)=(3p)^3-(10q)^3=27p^3-1000q^3\)

• \({\largeб)}\ (7m+2n)(4n^2-14mn+49m^2)=(7m+2n)(49m^2-14mn+4n^2)=(7m)^3+(2n)^3=343m^3+8n^3\)

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

• \({\largeг)}\ (km-n^2)(k^2m^2+kmn^2+n^4)=(km)^3-(n^2)^3=k^3m^3-n^6\)

• \({\largeд)}\ \left(4y^2-xy+\frac{1}{4}x^2\right)\left(\frac{1}{2}x+2y\right)=\left(\frac{1}{2}x+2y\right)\left(\frac{1}{4}x^2-xy+4y^2\right)=\left(\frac{1}{2}x\right)^3+(2y)^3=\frac{1}{8}x^3+8y^3\)

• \({\largeе)}\ (1{,}21q^2+0{,}22pq+0{,}04p^2)(0{,}2p-1{,}1q)=(0{,}2p-1{,}1q)(0{,}04p^2+0{,}22pq+1{,}21q^2)=(0{,}2p)^3-(1{,}1q)^3=0{,}008p^3-1{,}331q^3\)

• \({\largeж)}\ \left(\frac{1}{9}m^4+m^2nk+9n^2k^2\right)\left(\frac{1}{3}m^2-3nk\right)=\left(\frac{1}{3}m^2\right)^3-(3nk)^3=\frac{1}{27}m^6-27n^3k^3\)

• \({\largeз)}\ \left(1\frac{1}{2}a^3-0{,}5b^2\right)\left(2\frac{1}{4}a^6+\frac{3}{4}a^3b^2+0{,}25b^4\right)=\left(\frac{3}{2}a^3\right)^3-(0{,}5b^2)^3=\frac{27}{8}a^9-0{,}125b^6=3\frac{3}{8}a^9-0{,}125b^6\)