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

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

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Respuesta :

graceice graceice · Answer 1 · 2016-08-03T19:19:49+02:00

(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

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athleticregina athleticregina · Answer 2 · 2019-04-12T17:06:28+02:00

Answer:

Option (a) is correct.

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

Step-by-step explanation:

Given : Expression [tex]\left(y+3\right)\left(y^2-3y+9\right)[/tex]

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

Consider the given expression [tex]\left(y+3\right)\left(y^2-3y+9\right)[/tex]

Apply distributive law,

[tex](a+b)(c+d)=ac+ad+bc+bd[/tex], we have,

[tex]=yy^2+y\left(-3y\right)+y\cdot \:9+3y^2+3\left(-3y\right)+3\cdot \:9[/tex]

Simplify, we have,

[tex]=y^2y-3yy+9y+3y^2-3\cdot \:3y+3\cdot \:9[/tex]

Also,

[tex]=y^3-3y^2+9y+3y^2-9y+27[/tex]

Adding similar terms, we have,

[tex]=y^3+27[/tex]

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