IF X=2y+6 find the value of x³-8y³-36xy-216
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Answered by
502
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
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
10
Answer:
Step-by-step explanation:
X=2Y+6
X-2Y=6 CUBING BOTH SIDE (X-2Y)^3=6^3
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