Предмет: Алгебра, автор: malekseenko863

Разложить на множители (x-0,3z)³+(0,3z-2y)³+(2y-x)³.

ВладимирБ: (x-0.3z)³+(0.3z-2y)³+(2y-x)³=0.03(x-2y)(10x-3z)(20y-3z)

Ответы

Автор ответа: Universalka

2

$\displaystyle\bf\\(x-0,3z)^{3} +(0,3z-2y)^{3} +(2y-x)^{3}$

Сделаем замену :

$\displaystyle\bf\\x-0,3z=a \ \ ; \ \ 0,3z-2y=b \ \ ; \ \ 2y-x=c$

Заметим , что :

$\displaystyle\bf\\\underbrace{x-0,3z}_{a} + \underbrace{0,3z-2y}_{b}=\underbrace{x-2y}_{-c}\\\\\\\boxed{a+b=-c}\\\\\\a^{3} +b^{3} -(a+b)^{3} =a^{3} +b^{3} -a^{3} -b^{3} -3ab(a+b)=\\\\\\=-3ab\underbrace{(a+b)}_{-c}=3abc=\boxed{3(x-0,3z)(0,3z-2y)(2y-x)}$

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