To simplify the polynomial expression given, let's first rewrite it clearly. The expression appears to involve a product of two identical factors:
\[ (xy - 2)(x^2 + 1)(xy - 2)(x^2 + 1) \]
This can be simplified by recognizing that:
\[ ((xy - 2)(x^2 + 1))^2 \]
Next, we need to simplify the expression inside the parentheses:
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Expand \((xy - 2)(x^2 + 1)\): \[ (xy - 2)(x^2 + 1) = xy \cdot x^2 + xy \cdot 1 - 2 \cdot x^2 - 2 \cdot 1 \] \[ = x^3y + xy - 2x^2 - 2 \]
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Now square that result: \[ (x^3y + xy - 2x^2 - 2)^2 \] To do this, we can use the formula \((a + b + c + d)^2\): \[ = a^2 + b^2 + c^2 + d^2 + 2(ab + ac + ad + bc + bd + cd) \] where:
- \(a = x^3y\)
- \(b = xy\)
- \(c = -2x^2\)
- \(d = -2\)
Calculating the squares:
- \(a^2 = (x^3y)^2 = x^6y^2\)
- \(b^2 = (xy)^2 = x^2y^2\)
- \(c^2 = (-2x^2)^2 = 4x^4\)
- \(d^2 = (-2)^2 = 4\)
Calculating the cross products:
- \(ab = x^3y \cdot xy = x^4y^2\)
- \(ac = x^3y \cdot (-2x^2) = -2x^5y\)
- \(ad = x^3y \cdot (-2) = -2x^3y\)
- \(bc = xy \cdot (-2x^2) = -2x^3y\)
- \(bd = xy \cdot (-2) = -2xy\)
- \(cd = (-2x^2)(-2) = 4x^2\)
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Combine everything: After calculating all the terms, we group like terms:
\[ x^6y^2 + x^2y^2 + 4x^4 + 4 + 2(x^4y^2 - 2x^5y - 4x^3y - 2xy + 4x^2) \]
This expression can be rearranged by collecting terms together:
\[ = x^6y^2 + x^2y^2 + 4x^4 + 4 + 2x^4y^2 - 4x^5y - 8x^3y - 4xy + 8x^2 \]
Finally, combining like terms will give us the fully simplified expression.
Alternatively, based on your question context, the second part you mentioned, x^3y + x^3y + x^2 + x^2 + xy + xy, simplifies quite easily:
\[ 2x^3y + 2x^2 + 2xy = 2(x^3y + x^2 + xy) \]
The overall simplifications yield that the complete result would be a combination accordingly to combine with the coefficient scale.
Please confirm any specific paths or the exact elements you want further detailed or reanalyzed in polynomial simplification.
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