Представьте в виде многочлена: а) (3x² − 1)(3x² + 1); б) (5a − b³)(b³ + 5a); в) (3/7m³ + 1/4n³)(3/7m³ − 1/4n³); г) (1/15 − 1/8p⁶)(1/8p⁶ + 1/15); д) (0,4y³ + 5a²)(5a² − 0,4y³); е) (1,2c² − 7a²)(1,2c² + 7a²); ж) (5/8x + y⁵)(y⁵ − 5/8x); з) (1/7p⁵ − 0,01)(0,01 + 1/7p⁵).

Используем формулу разности квадратов: \((u-v)(u+v)=u^2-v^2\).

1. \((3x^2-1)(3x^2+1)=(3x^2)^2-1^2=9x^4-1\).
2. \((5a-b^3)(b^3+5a)=(5a-b^3)(5a+b^3)=25a^2-b^6\).
3. \(\left(\frac37m^3+\frac14n^3\right)\left(\frac37m^3-\frac14n^3\right)=\frac9{49}m^6-\frac1{16}n^6\).
4. \(\left(\frac1{15}-\frac18p^6\right)\left(\frac18p^6+\frac1{15}\right)=\frac1{225}-\frac1{64}p^{12}\).
5. \((0.4y^3+5a^2)(5a^2-0.4y^3)=(5a^2)^2-(0.4y^3)^2=25a^4-0.16y^6\).
6. \((1.2c^2-7a^2)(1.2c^2+7a^2)=(1.2c^2)^2-(7a^2)^2=1.44c^4-49a^4\).
7. \(\left(\frac58x+y^5\right)\left(y^5-\frac58x\right)=y^{10}-\frac{25}{64}x^2\).
8. \(\left(\frac17p^5-0.01\right)\left(0.01+\frac17p^5\right)=\frac1{49}p^{10}-0.0001\).

Ответ: \(9x^4-1;\ 25a^2-b^6;\ \frac9{49}m^6-\frac1{16}n^6;\ \frac1{225}-\frac1{64}p^{12};\ 25a^4-0.16y^6;\ 1.44c^4-49a^4;\ y^{10}-\frac{25}{64}x^2;\ \frac1{49}p^{10}-0.0001\).