LHS = (y–3)^2, which is similar to (a–b)^2identity, where (a–b)^2= a^2+b^2-2ab. (y–3)^2= y^2+(3)^2–2y×3 = y^2+9–6y ≠y^2–9 = RHS The correct statement is (y–3)^2= y^2+ 9–6y
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