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

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

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

(1𝑦/3+3)^3

(𝑦/3+3)^3

(𝑦/3+3⋅3/3)^3

(𝑦+3⋅3/3)^3

(𝑦+9/3)^3

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

a: \(\left(6x^2+\frac13\right)^2\)

\(=\left(6x^2\right)^2+2\cdot6x^2\cdot\frac13+\left(\frac13\right)^2\)

\(=36x^4+4x^2+\frac19\)

b: \(\left(5x-4y\right)^2=\left(5x\right)^2-2\cdot5x\cdot4y+\left(4y\right)^2\)

\(=25x^2-40xy+16y^2\)

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) \(=8x^3+y^3\)

\(=4x^2y^2-12x^3y^4+9x^2y^6\)

a,(x+2y)³ =x³+3.x².2y+3.x.(2y)²+(2y)³

= x³+6x²y+12xy²+8y³

b, phần b tương tự dấu thay đổi một tí

c, (5x+1)(5x+1)= (5x+1)²

=25x²+10x+1

a)  \(\left(x+2y\right)^3=x^3+6x^2y+12xy^2+8y^3\)

b)  \(\left(2x-1\right)^3=8x^3-12x^2+6x-1\)

c)  \(\left(5x+1\right)\left(5x-1\right)=25x^2-1\)