\(\left(\dfrac{1}{3y+3}\right)^3=\dfrac{1}{\left(3y+3\right)^3}=\dfrac{1}{27y^3+81y^2+81y+27}\)

\(\left(\dfrac{1}{3y+3}\right)^3=\dfrac{1^3}{\left(3y+3\right)^3}=\dfrac{1}{27\left(y^3+3y^2+3y+1\right)}\)

\(\left(\dfrac{1}{3}y+3\right)^3=\dfrac{1}{27}y^3+y^2+9y+27\)

a: Ta có: \(\left(x+y\right)^3-\left(x-y\right)^3\)

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

\(=6x^2y+2y^3\)

\(\left(x+y\right)^3-\left(x-y\right)^3\)

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

\(=8y^3+6y\left(x^2-y^2\right)\)

\(=8y^3+6x^2y-6y^3\)

\(=2y^3+6x^2y\)

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

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

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

a: Ta có: \(\left(x+y\right)^3-\left(x-y\right)^3\)

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

\(=6x^2y+2y^3\)

Ta có:(a²+ab+b²)(a²-ab+b²)-(a⁴+b⁴)

    =   (a²+b²)²-a²b²-a⁴-b⁴=a⁴+2a²b²+b⁴-a²b²-a⁴-b⁴=a²b²

Ta có:(a²+ab+b²)(a²-ab+b²)-(a⁴+b⁴)

    =   (a²+b²)²-a²b²-a⁴-b⁴=a⁴+2a²b²+b⁴-a²b²-a⁴-b⁴=a²b²

a) $(3x+5)^2\\=(3x)^2+2.3x.5+5^2\\=9x^2+30x+25$

b) $(6x+\dfrac{1}{3})^2\\=(6x)^2+2.6x.\dfrac{1}{3}+(\dfrac{1}{3})^2\\=36x^2+4x+\dfrac{1}{9}$

c) $(5x-4y)^2\\=(5x)^2-2.5x.4y+(4y)^2\\=25x^2-40xy+16y^2$

d) $(5x-3)(5x+3)\\=(5x)^2-(3)^2\\=25x^2-9$

a) \(=x^3+27-54-x^3=-27\)

b) \(=8x^3+y^3\)

Lời giải:

\(A=(x+2y+3z)(x-2y+3z)\)

\(=[(x+3z)+2y][(x+3z)-2y]\)

\(=(x+3z)^2-(2y)^2\)

\(=x^2+9z^2+6xz-4y^2\)

\(7)\)  \(\left(3x\right)^2-y^2=\left(3x-y\right)\left(3x+y\right)\)

\(8)\)  \(x^2-\left(2y\right)^2=\left(x-2y\right)\left(x+2y\right)\)

\(9)\)  \(\left(2x\right)^2-y^2=\left(2x-y\right)\left(2x+y\right)\)