2x^2+4xy+4y^2=0
where (x=X+3)
(y=Y+4)
SUBSTITUTE IT
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Answered by
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Answer:
2x^3+4xy+4y^2 = 2(X+3)^2+4(X+3)(Y+4)+4(Y+4)^2
=2(X^2+6X+9)+4(XY+4X+3Y+12)+4(Y^2+8Y+16)
= 2x^2+12X+18+4XY+16X+12Y+48+4Y^2+32Y+64
=2x^2+4Y^2+4XY+12X+16X+12Y+32Y+18+48+64
=2x^2+4Y^2+4XY+28X+44Y+130
Step-by-step explanation:
Answered by
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Answer:
Mark as brainliest answer.
Step-by-step explanation:
→ 2x² + 4xy + 4y² = 0
→ x = x+3
→ y = y+4
→ 2(x+3)² + 4(x+3)(y+4) + 4(y+4)² = 0
→ 2(x² + 9 + 6x) + 4(xy + 4x + 3y + 12) + 4(y² + 16 + 8y)
→ 2x² + 18 + 12x + 4xy + 16x + 12y + 48 + 4y² + 64 + 32y
→ 2x² + 12x + 16x + 12y + 32y + 4xy + 4y² + 18 + 48 + 64
→ 2x² + 28x + 44y + 4xy + 4y² + 130
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