Question

(1/3x+2).(3x-6) 

(x^2-3x+9).(x+3)

(-2xy+3).(xy-1)

x(xy-1).(xy+1)

Answer 1

\(\left(\frac{1}{3}x+2\right)\left(3x-6\right)\)

\(=\frac{1}{3}x\left(3x-6\right)+2\left(3x-6\right)\)

\(=x^2-2x+6x-12\)

\(=x^2+4x-12\)

\(\left(x^2-3x+9\right)\left(x+3\right)\)

\(=\left(x+3\right)\left(x^2-3x+9\right)\)

\(=x^3+27\)

Answer 2

a) (1/3x+2).(3x-6) = (1​​/3x.3x)+(1/3x.-6)+(2.3x)+(2.-6) = x^2-2x+6x-12 = x^2+4x-12 = x^2+2x2+2^2-16 = (x+2)^2-16

b) (x^2-3x+9).(x+3) = (x^2.x)+(x^2.3)+(-3x.x)+(-3x.3)+(9.x)+(9.3) = x^3+3x^2-3x^2-9x+9x+27 = x^3+27

c) (-2xy+3).(xy-1) = (-2xy.xy)+(-2xy.-1)+(3.xy)+(3.-1) = -2x^2y^2+2xy+3xy-3 = -2x^2y^2+5xy-3

d) x(xy-1).(xy+1) = (x^2y-x).(xy+1) = (x^2y.xy)+(x^2y.1)+(-x.xy)+(-x.1) = x^3y^2+x^2y-x^2y-x = x^3y^2-x