Question

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

Answer 1

`a)(x+3)(x^2-3x+9)`

`=(x. x^2)-(x.3x)+(x.9)+(3.x^2)-(3.3x)+(3.9)`

`=x^3-3x^2+9x+3x^2-9x+27`

`=x^3+(-3x^2+3x^2)+(9x-9x)+27`
`=x^3+27`

 

 

Answer 2

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

= \(\left(x\cdot x^2\right)-\left(x\cdot3x\right)+\left(x\cdot9\right)+\left(3\cdot x^2\right)-\left(3\cdot3x\right)+\left(3\cdot9\right)\)

= \(x^3-3x^2+9x+3x^2-9x+27\)

= \(x^3+\left(-3x^2+3x^2\right)+\left(9x-9x\right)+27\)

= \(x^3+27\)

Answer 3

`a)(x+3)(x^2-3x+9)`

`=x(x^2-3x+9)+3(x^2-3x+9)`

`=x^3-3x^2+9x+3x^2-9x+27`

`=x^3+(-3x^2+3x^2)+(9x-9x)+27`

`=x^3+27`