a) \(\left(3x-5\right)\left(3x+5\right)\)

\(=\left(3x\right)^2-5^2\)

\(=9x^2-25\)

b) \(\left(x-2y\right)\left(x+2y\right)\)

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

\(=x^2-4y^2\)

c) \(\left(-x-\dfrac{1}{2}y\right)\left(-x+\dfrac{1}{2}y\right)\)

\(=\left(-x\right)^2-\left(\dfrac{1}{2}y\right)^2\)

\(=x^2-\dfrac{1}{4}y^2\)

`a, (3x-5)(3x+5) = 9x^2 - 25`

`b, (x-2y)(x+2y) = x^2 -4y^2`

`c, (-x-1/2y)(-x+1/2y) = x^2 - 1/4y^2`

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\(a.10x\left(x-y\right)-6y\left(y-x\right)\\ =10x\left(x-y\right)+6y\left(x-y\right)\\ =\left(10x-6y\right)\left(x-y\right)\\ =2\left(5x-3y\right)\left(x-y\right)\)

\(b.14x^2y-21xy^2+28x^3y^2\\ =7xy\left(x-y+xy\right)\)

\(c.x^2-4+\left(x-2\right)^2\\ =\left(x-2\right)\left(x+2\right)+\left(x-2\right)^2\\ =\left(x-2\right)\left(x+2+x-2\right)\\ =2x\left(x-2\right)\)

\(d.\left(x+1\right)^2-25\\ =\left(x+1-5\right)\left(x+1+5\right)=\left(x-4\right)\left(x+6\right)\)

`a, (a-1)(a+1)(a^2+1)`

`= (a^2-1)(a^2+1)`

`= a^4-1`

`b, (xy+1)^2 - (xy-1)^2`

`= x^2y^2 + 2xy + 1 - x^2y^2 + 2xy - 1`

`= 4xy`

a) \(\left(a-1\right)\left(a+1\right)\left(a^2+1\right)\)

\(=\left(a^2-1\right)\left(a^2+1\right)\)

\(=a^4-1\)

b) \(\left(xy+1\right)^2-\left(xy-1\right)^2\)

\(=\left[\left(xy+1\right)-\left(xy-1\right)\right]\left[\left(xy+1\right)+\left(xy-1\right)\right]\)

\(=\left(xy+1-xy+1\right)\left(xy+1+xy-1\right)\)

\(=4xy\)

\(\begin{array}{l}\left( {x - 2y} \right)\left( {{x^2} + 2xy + 4{y^2}} \right) + \left( {x + 2y} \right)\left( {{x^2} - 2xy + 4{y^2}} \right)\\ = {x^3} - {\left( {2y} \right)^3} + {x^3} + {\left( {2y} \right)^3}\\ = {x^3} - 8{y^3} + {x^3} + 8{y^3}\\ = 2{x^3}\end{array}\)

`a, (2x-3)^3 = 8x^3 - 36x^2 + 54x - 27`

`b, (a+3b)^3 = a^3 + 9a^2b + 27ab^2 + 27b^3`

`c, (xy-1)^3 = x^3y^3 - 3x^2y^2 + 3xy -1`

1)  2xy²+x²y⁴+1=(xy²)²+2xy².1+1²=(xy²+1)²

2)

a)2(x-y)(x+y)+(x+y)²+(x-y)²=(x+y+x-y)²=(2x)²=4x²

b)(x-y+z)²+(z-y)²+2(x-y+z)(y-z)

=(x-y+z)²+(y-z)²+2(x-y+z)(y-z)

=(x-y+z+y-z)²

=x²