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i) x³ + y³ = (x + y)(x² - xy + y² )

Taking the RHS,

(x + y)(x² - xy + y²) = x(x² - xy + y² ) + y(x² - xy + y²)

= x³ - x²y + xy² + x²y - xy² + y³

= x³ + y³ = LHS

ii) x³ - y³ = (x - y)(x² + xy + y²)

Taking the RHS,

(x - y)(x² + xy + y²) = x (x² + xy + y²) - y (x² + xy + y²)

= x³ + x²y + xy² - x²y - xy² - y³

= x³ - y³ = LHS

in this question we will use the identity 

(ax + b) * (cx + d) = ac*x^2 + (ad + bc)* x + bd

(1) (x + 4) (x + 10) = x^2 + (4 + 10)*x + 4 * 10

                               =x^2 + 14x + 40

(2) (x +8) (x - 10) = x^2 + (8 - 10)x + 8 * -10

                             = x^2 - 2x - 80

(3)(3x + 4) (3x - 5) = (3x)^2 + (4*3 + (- 5)*3) * x + 4 * (-5)

                              = 9x^2 - 3x - 20

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

(a + b)(a - b) = a² - b²

(i) 9x² + 6xy + y²

= (3x)² + 2(3x)( y) + (y)²

Identity: (a + b)² = a² + 2ab + b²

Considering a = 3x and b = y 

Hence 9x² + 6xy + y² = (3x + y)²

(ii) 4y² - 4y + 1

= (2y)² - 2(2y)(1) + (1)²

Identity: (a - b)² = a² - 2ab + b²

Considering a = 2y and b = 1

Hence 4y² - 4y + 1 = (2y - 1)²

(iii) x² - y²/100

= x² - (y/10)²

Identity: (a + b)(a - b) = a² - b²

Considering a = x and b = y/10

Hence, x² - y²/100 = (x + y/10)(x - y/10)

Using $(A+B+C)^2=A^2+B^2+C^2+2AB+2AC+2BC$:

(i) $(x+2y+4z)^2=x^2+4y^2+16z^2+4xy+8xz+16yz$

(ii) $(2x-y+z)^2=4x^2+y^2+z^2-4xy+4xz-2yz$

(iii) $(-2x+3y+2z)^2=4x^2+9y^2+4z^2-12xy-8xz+12yz$

(iv) $(3a-7b-c)^2=9a^2+49b^2+c^2-42ab-6ac+14bc$

(v) $(-2x+5y-3z)^2=4x^2+25y^2+9z^2-20xy+12xz-30yz$

(vi) $\left(\tfrac14 a-\tfrac12 b+1\right)^2=\tfrac1{16}a^2+\tfrac14 b^2+1-\tfrac14 ab+\tfrac12 a-b$