c: Ta có: \(x^3-12x^2+48x-64=0\)

\(\Leftrightarrow x-4=0\)

hay x=4

c: Ta có: \(x^3-12x^2+48x-64=0\)

\(\Leftrightarrow x-4=0\)

hay x=4

\(d,\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=15\\ \Leftrightarrow24x=-10\Leftrightarrow x=-\dfrac{5}{12}\\ e,\Leftrightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x=17\\ \Leftrightarrow9x=10\Leftrightarrow x=\dfrac{10}{9}\\ f,\Leftrightarrow9x^2+18x+9-18x=36+x^3-27\\ \Leftrightarrow x^3-9x^2=0\Leftrightarrow x^2\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\)

a)
\(=x^3+3.x^2.1+3.x.1^2+1^3\)
\(=x^3+3x^2+3x+1\)
b)
\(=\left(2x\right)^3+3.\left(2x\right)^2.3+3.2x.3^2+3^3\)
\(=8x^3+36x^2+54x+27\)
c)
\(x^3+3.x^2.\dfrac{1}{2}+3.x.\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3\)
\(=x^3+1,5x^2+0,75x+0,125\)

d)
=\(\left(x^2\right)^3-3.\left(x^2\right)^2.2+3.x^2.2^2-2^3\)
\(=x^5-6x^4+12x^2-8\)
e)
\(=\left(2x\right)^3-3.\left(2x\right)^2.3y+3.2x.\left(3y\right)^2-\left(3y\right)^3\)
\(=8x^3-36x^2y+54xy^2-27y^3\)

\(D=x^3-3x^2y+3xy^2-y^3-3x^3+6x^2y-3xy^2+3x^3-3x^2y-x^3\\ D=-y^3\)

\(6,\Rightarrow6x^2+12x+6-2x^3-6x^2-6x-2+2x^3-2=1\\ \Rightarrow6x=-1\Rightarrow x=-\dfrac{1}{6}\\ 7,\Rightarrow\left(x-2\right)\left(x-2-x-2\right)=0\\ \Rightarrow-4\left(x-2\right)=0\\ \Rightarrow x-2=0\Rightarrow x=2\\ 8,\Rightarrow\left(x-2\right)^2-25=0\\ \Rightarrow\left(x-2-5\right)\left(x-2+5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)

Lời giải:

 Áp dụng HĐT: $(a-b)^3=a^3-b^3-3ab(a-b)$ cho cả hai bạn.

a. 

$M=x^3-1-3x(x-1)-3x(x-1)^2+3x^2(x-1)+x^3$
$=2x^3-1+3x(x-1)[-1-(x-1)+x]$

$=2x^3-1+3x(x-1).0=2x^3-1$

b.

$D=[(x-y)-x]^3=-y^3$