Math, asked by akshatmangal101, 8 months ago

Find the value of x³ - 8y³ - 36xy - 216 , when x = 2y + 6​

Answers

Answered by SuraBhaavya
3

Step-by-step explanation:

X=2y+6 

x^3-8y^3-36xy-216 

=(2y+6)³−8y³−36(2y+6)y−216 

=8y³+6³ +3×(2y)²×6+3×2y×6² −8y³−72y²−216y−216 

=8y³+216 +3×4y²×6+3×2y×36−8y³ −72y²−216y−216 

=8y³+216 +72y²+216y−8y³ −72y²−216y−216 

=8y³−8y³ +72y²−72y²+216y −216y+216−216 

=0

Answered by tennetiraj86
1

Answer:

answer for the given problem is given

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