Question

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

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

Answer:

Option (a) is correct.

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

Step-by-step 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 $y^3+27$