a: \(\left(3x-2\right)^2=9x^2-12x+4\)

c: \(9x^2-225=9\left(x^2-25\right)=9\left(x-5\right)\left(x+5\right)\)

a: \(\left(3x-2\right)^2=9x^2-12x+4\)

c: \(9x^2-225=\left(3x-15\right)\left(3x+15\right)\)

d: \(\left(2x-3y\right)^3=8x^3-36x^2y+54xy^2-27y^3\)

a: \(\left(\dfrac{1}{3}x+2y\right)\left(\dfrac{1}{9}x^2-\dfrac{2}{3}xy+4y^2\right)=\dfrac{1}{27}x^3+8y^3\)

b: \(\left(x^2-\dfrac{1}{3}\right)\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)=x^6-\dfrac{1}{27}\)

c: \(\left(y-5\right)\left(y^2+5y+25\right)=y^3-125\)

a) \(\left(a+b\right)^3-\left(a-b\right)^3=a^3+3a^2b+3ab^2+b^3-\left(a^3-3a^2b+3ab^2-b^3\right)=6a^2b+b^3=b\left(6a^2+b^2\right)\)

b)  \(\left(x+y\right)^3+\left(x-y\right)^3=\left(x^3+3x^2y+3xy^2+y^3\right)+\left(x^3-3x^2y+3xy^2-y^3\right)=2x^3+6xy^2=2x\left(x^2+3y^2\right)\)

a) \(=\left(a+b-a+b\right)\left[\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)

\(=2b\left(a^2+2ab+b^2+a^2-b^2+a^2-2ab+b^2\right)\)

\(=2b\left(3a^2+b^2\right)\)

b) \(=\left(x+y+x-y\right)\left[\left(x+y\right)^2-\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)

\(=2x\left(x^2+2xy+y^2-x^2+y^2+x^2-2xy+y^2\right)\)

\(=2x\left(x^2+3y^2\right)\)

a: \(\left(a+b\right)^3-\left(a-b\right)^3\)

\(=a^3+3a^2b+3ab^2+b^3-a^3+3a^2b-3a^2b+b^3\)

\(=6a^2b+2b^3\)

\(=2b\left(3a^2+b^2\right)\)

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.

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

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