Question

What is the product of (y+3)(y^2-3y+9)

y3 + 27
y3 – 27
y3 – 6y2 + 27
y3 + 6y2 + 27

Answer 1

(y+3)(y^2-3y+9)
=y(y^2-3y+9)
+3(y^2-3y+9)
=(y^3-3y^2+9y)+(3y^2-9y+27)
=y^3-3y^2+9y+3y^2-9y+27
=y^3+27

Answer 2

Option (a) is correct.

The product of the given expression
`\left(y+3\right)\left(y^2-3y+9\right)` is
`y^3+27`

Explanation:

Given : Expression
`\left(y+3\right)\left(y^2-3y+9\right)`

We have to find the product of the given expression
`\left(y+3\right)\left(y^2-3y+9\right)`

Consider the given expression
`\left(y+3\right)\left(y^2-3y+9\right)`

Apply distributive law,

`(a+b)(c+d)=ac+ad+bc+bd`, we have,

`=yy^2+y\left(-3y\right)+y\cdot \:9+3y^2+3\left(-3y\right)+3\cdot \:9`

Simplify, we have,

`=y^2y-3yy+9y+3y^2-3\cdot \:3y+3\cdot \:9`

Also,

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

Adding similar terms, we have,

`=y^3+27`

Thus, The product of the given expression
`\left(y+3\right)\left(y^2-3y+9\right)` is
`y^3+27`