Algebraic identities

\(x^2+2xy+y^2=(x+y)^2\), so it is a perfect-square trinomial. Check the middle term: \(2xy=2\cdot x\cdot y\). In option B, the middle term is only \(xy\), so it cannot be \((x+y)^2\). Exam tip: compare the middle term with twice the product of the square roots of the end terms.

Since \(81m^4=(9m^2)^2\) and \(16n^6=(4n^3)^2\), option A has the form \(a^2-b^2\) and factors as \((9m^2-4n^3)(9m^2+4n^3)\). Option B is a sum, not a difference. Exam tip: check for two square terms separated by a minus sign.

In the cube of a difference, signs occur in (+), (-), (+), (-) order. Exam tip: check both (3^2) and (3^3).

( (2x)^3=8x^3 ), (3(2x)^2\cdot3=36x^2), and (3(2x)\cdot9=54x). Exam tip: check the power of every term.

\(9x^2-24xy+16y^2=(3x-4y)^2\), so it has two identical binomial factors. Its middle term is \(-2(3x)(4y)=-24xy\). In option B, the last term would need to be \(16y^2\). Exam tip: check square roots of the first and last terms.

The expression is a difference of two cubes because \\(125p^3=(5p)^3\\) and \\(8q^3=(2q)^3\\). The identity for a difference of cubes is \\(a^3-b^3=(a-b)(a^2+ab+b^2)\\). Substituting \\(a=5p\\) and \\(b=2q\\) gives \\((5p-2q)(25p^2+10pq+4q^2)\\).

Thus option A is correct. The second factor contains a plus sign between all three terms; this is a key feature of the difference-of-cubes identity. Option B uses the corresponding sum-of-cubes pattern, while C uses incorrect cube roots and D treats the whole expression as the cube of a binomial. Expansion of option A returns the original expression.

The cube identity is \((a+b)^3=a^3+3a^2b+3ab^2+b^3\), so A is correct. In B, the sign of the last term is wrong. Exam tip: remember the coefficient pattern 1, 3, 3, 1.

The end terms have roots \(2p\) and \(-3q\); twice their product is \(-12pq\). Therefore, A is \((2p-3q)^2\). C fails the middle-term test. Exam tip: always verify the middle coefficient.

\(4x^2-12xy+9y^2=(2x)^2-2(2x)(3y)+(3y)^2=(2x-3y)^2\), so A is a perfect square. In B, the middle term should be \(-12xy\), not \(-10xy\). Exam tip: check the middle term using twice the product of square roots.

(81=35+2(ab+bc+ca)), so (2(ab+bc+ca)=46) and the value is (23). Exam tip: find the double product sum first.

The (xy) term is (12xy), the (yz) term is (6yz), and the (xz) term is (4xz). Exam tip: find each double product separately.

Most errors occur due to incomplete terms, wrong signs, or wrong powers. Exam tip: identify the whole term before applying the formula.

\((u-v)^2=u^2-2uv+v^2\), so this trinomial is the square of a difference. In contrast, \((u+v)^2\) has the middle term \(+2uv\). Exam tip: always check the sign of the middle term.

In \((a+b+c)^2\), all three terms are positive, so the pairwise products \(2ab\), \(2bc\), and \(2ca\) are all positive. A negative sign would make the related cross-term negative. Exam tip: check signs carefully in squared expansions.

Twice the product of (-2b) and (c) is (2\cdot(-2b)\cdot c=-4bc). Match both sign and coefficient carefully.

\(x^2+10x+25=x^2+2\cdot x\cdot5+5^2=(x+5)^2\), so it is a perfect square of a binomial. In option B, the constant term should be \(5^2=25\), not 20. Exam tip: match the middle term with \(2ab\).

\(4x^2-25=(2x)^2-5^2\), so it is a difference of two perfect squares and factors as \((2x-5)(2x+5)\). Options B and D are perfect-square trinomials. Exam tip: check for two square terms separated by a minus sign.

\((a+b)^2=a^2+2ab+b^2\), so its middle term is \(+2ab\). In contrast, \((a-b)^2\) has \(-2ab\) as its middle term. Exam tip: check the sign between the two terms before choosing a square identity.

Here \(16a^2=(4a)^2\) and \(25b^2=(5b)^2\), while the middle term is \(-2\times4a\times5b=-40ab\). Hence it is \((4a-5b)^2\). Exam tip: always verify the sign of the middle term.

\((1000+3)^2=1000^2+2(1000)(3)+3^2=1000000+6000+9=1006009\). The student omitted the middle term \(2ab\). In exams, write all three terms of \((a+b)^2\).