Question

№34.1(чет)№34.3(чет),№34.7(1,3)
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Answer 1

№34.1 (2,4,6,8)

$y^3-b^3=(y-b)(y^2+yb+b^2)\\\\m^3+k^3=(m+k)(m^2-mk+k^2)\\\\64+s^3=4^3+s^3=(4+s)(4^2-4s+s^2)=(4+s)(16-4s+s^2)\\\\1000+y^3=10^3+y^3=(10+y)(10^2-10y+y^2)=(10+y)(100-10y+y^2)$

№34.3 (2,4,6,8)

$\frac{1}{27}+c^3=(\frac{1}{3})^3+c^3 =(\frac{1}{3}+c)((\frac{1}{3})^2-\frac{1}{3}c+c^2)=(\frac{1}{3}+c)(\frac{1}{9}-\frac{1}{3}c+c^2) \\\\\frac{1}{125}+t^3=(\frac{1}{5})^3+t^3 =(\frac{1}{5}+t)((\frac{1}{5})^2-\frac{1}{5}t+t^2)=(\frac{1}{3}+t)(\frac{1}{25}-\frac{1}{5}t+t^2)\\\\y^3-\frac{27}{64}=y^3-(\frac{3}{4})^3 =(y^3-\frac{3}{4})(y^2+\frac{3}{4}y+(\frac{3}{4})^2)=(y^3-\frac{3}{4})(y^2+\frac{3}{4}y+\frac{9}{16})\\$

$\frac{1}{216}-z^3=(\frac{1}{6})^3-z^3 =(\frac{1}{6}-z)((\frac{1}{6})^2+\frac{1}{6}z+z^2)=(\frac{1}{36}+z)(\frac{1}{6}+\frac{1}{6}z+z^2)$

№34.7 (1)

$(2x-3)(4x^2+6x+9)-8x^3=2,7x\\8x^3-27-8x^3=2,7x\\-27=2,7x\\2,7x=-27\\x=-27:(2,7)\\x=-10$

Решено по формуле  суммы кубов ⇒  a³ + b³ = ( a + b )( a² - ab + b² )

и формуле разности кубов ⇒  a³ - b³ = ( a - b )( a² + ab + b² )