Algebraic identities

\(a^2+2ab+b^2=(a+b)^2\), so it is always a perfect-square trinomial. Option B has middle term \(ab\), but a square of a binomial requires \(2ab\). Exam tip: check the middle coefficient first.

By identity the left side is (4\cdot3m\cdot n=12mn). (12mn=48mn) gives no valid nonzero fixed (m); the given equality is not generally true.

Expand each product: \((x+4)(x-6)=x^2-2x-24\) and \((x-4)(x+6)=x^2+2x-24\). On adding, the \(-2x\) and \(+2x\) terms cancel, giving \(2x^2-48\). In option B, the sign of the constant term is incorrect. Exam tip: after expanding, check the signs carefully before combining like terms.

\((x+4)(x-6)=x^2-2x-24\), while \((x-4)(x+6)=x^2+2x-24\). On subtracting the second expression, \(x^2-2x-24-(x^2+2x-24)=-4x\). Therefore, the correct answer is \(-4x\). The option \(4x\) results from an incorrect sign while subtracting the bracketed expression. Exam tip: when subtracting a whole expression, change the sign of every term inside its bracket.

In three linear factors, the coefficient of (x^2) is the sum of constants. Here (-2+5-7=-4).

The constant term is (1\cdot(-3)\cdot8=-24). Exam tip: multiply only the constants for constant term.

Here \(27=3^3\), so \(x^3+27=x^3+3^3=(x+3)(x^2-3x+9)\), which fits the sum-of-cubes identity. \(x^3-27\) is a difference of cubes. Exam tip: check whether the constant is a perfect cube.

\(a^2-b^2\) is the difference of two perfect squares, so it factorises as \((a+b)(a-b)\). Option B is the identity for the square of a binomial, not a difference of squares. Exam tip: identify both square terms first.

\((a+b)^2=(a+b)(a+b)\). Multiplying gives \(a^2+ab+ab+b^2=a^2+2ab+b^2\). Option B is the expansion of \((a-b)^2\). Exam tip: remember that the middle term in a square expansion has coefficient 2.

The correct identity is \((x+y)(x-y)=x^2-y^2\). On multiplying, the middle terms \(-xy\) and \(+xy\) cancel, leaving a difference of squares. Exam tip: look for binomials with opposite signs.

In \((a+b)(a-b)\), the middle terms \(-ab\) and \(+ab\) cancel, leaving \(a^2-b^2\). In \((a+b)^2\), a \(+2ab\) term appears. Exam tip: identify same binomials with opposite signs as a difference of squares.

The expression has the form \(a^2-b^2\), so the difference-of-squares identity is the quickest method: \\(a^2-b^2=(a+b)(a-b)\\). Take \(a=1004\) and \(b=996\). Their sum is 2000, and their difference is 8. Therefore the required value is \\(2000\times8=16000\\).

Hence option A is correct. Calculating both large squares separately is unnecessary and creates more opportunity for arithmetic mistakes. The value 8000 would result from losing a factor of 2, while 4000 and 12000 do not match the product of the sum and difference. The identity works because the middle terms cancel when the two squares are subtracted.

Here, 503 = 500 + 3 and 497 = 500 - 3. Using the identity \((a+b)^2+(a-b)^2=2(a^2+b^2)\), we get \((503^2+497^2)=2(500^2+3^2)=2(250000+9)=500018\). The value 500000 includes only \(2\times500^2\); the term \(2\times3^2\) must also be added. Exam tip: When two numbers are equally spaced from a common middle number, use this identity for quick calculation.

By difference of squares, ( (203+197)(203-197)=400\cdot6=2400 ). Exam tip: directly find sum and difference.

Difference of squares gives ( (x^2+1+x^2)(x^2+1-x^2)=2x^2+1 ). Exam tip: simplify at the end.

\(9p^2-24pq+16q^2=(3p)^2-2(3p)(4q)+(4q)^2=(3p-4q)^2\), so A is a perfect-square trinomial. In B, the last term is not \((4q)^2\). Exam tip: match the middle term with \(-2ab\).

The constant term is (1\cdot2\cdot3=6). Exam tip: multiply only constants for the constant term.

To find the constant term, take the term without \(x\) from each factor. Thus, the constant term is \(1\times(-3)\times5=-15\). The closest distractor, \(15\), results from ignoring the negative sign. Exam tip: multiply only the constant terms and track signs carefully.

With \(x=a,\ y=3b\), \((x+y)^2\) has middle term \(2(a)(3b)=6ab\) and last term \(9b^2\). So A is correct; D represents \((a-3b)^2\). Exam tip: check the middle-term sign.

( (u+v)^2-(u-v)^2=4uv ). With (u=4x) and (v=y), it gives (16xy), so (32xy) is not correct.