Algebraic identities

The expression is a difference of two squares. The identity \\(a^2-b^2=4ab\\) is not the general identity; rather, when the squares are \\((a+b)^2-(a-b)^2\\), their difference is \\(4ab\\). Here, take \\(a=6x\\) and \\(b=1\\). Therefore, \\((6x+1)^2-(6x-1)^2=4(6x)(1)=24x\\). Thus, option B is correct.

Expanding also confirms the result. We have \\((6x+1)^2=36x^2+12x+1\\) and \\((6x-1)^2=36x^2-12x+1\\). Subtracting the second expression cancels both \\(36x^2\\) and the constant terms, while the linear terms give \\(12x-(-12x)=24x\\). Option A, \\(12x\\), misses half of the difference. Options C and D incorrectly retain square or constant terms that cancel. The complete term \\(6x\\), not merely 6, must be used for a.

The identity is \(a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\), so A is correct. In B, the signs of the mixed terms are wrong. Exam tip: first check the factor \(a+b+c\).

\(x^2-49=x^2-7^2\) is a difference of two squares. Using \(a^2-b^2=(a-b)(a+b)\), it becomes \((x-7)(x+7)\). \(x^2+49\) is a sum of squares, not this identity. Exam tip: check for two perfect squares separated by a minus sign.

To find the coefficient of the term containing \\(a^2b\\) in \\((3a-b)^3\\), use the cube-of-a-difference identity \\((x-y)^3=x^3-3x^2y+3xy^2-y^3\\). Here, take \\(x=3a\\) and \\(y=b\\). The second term is \\(-3(3a)^2b\\), which becomes \\(-3\cdot9a^2b=-27a^2b\\). Thus the required coefficient is \\(-27\\), including its negative sign.

Hence option B follows. A common error is to calculate only \\(3(3a)^2b=27a^2b\\) and forget that the middle term has a minus sign. Option A gives the first term’s coefficient, not the coefficient of \\(a^2b\\). Options C and D result from incorrect multiplication or an incorrect sign. The complete expansion confirms the answer.

Use the cube identity \\(a+b)^3=a^3+3a^2b+3ab^2+b^3\\) and \\(a-b)^3=a^3-3a^2b+3ab^2-b^3\\). Put \\(a=x\\) and \\(b=4\\). Their difference is \\(6a^2b+2b^3\\), so it becomes \\(6x^2(4)+2(4^3)=24x^2+128\\). The \\(x^3\\) terms cancel, and the terms containing only \\(x\\) also cancel.

Thus option A is correct. Direct expansion gives \\(x^3+12x^2+48x+64\\) minus \\(x^3-12x^2+48x-64\\), which leaves \\(24x^2+128\\). Option B has only half of the quadratic contribution, option C incorrectly leaves an \\(x^3\\) term, and option D includes a linear term that should cancel.

In \(x^2+10x+25\), the first and last terms are \(x^2\) and \(5^2\). The middle term is \(2\times x\times5=10x\), so it equals \((x+5)^2\). Exam tip: check whether the middle term is \(2ab\).

This is ( (4x)^3-(3y)^3 ). Exam tip: in difference of cubes, all terms of the second factor are positive.

The governing concept is the cube identity followed by extraction of a common factor. Expand the binomial: (a+b)^3=a^3+3a^2b+3ab^2+b^3. Now subtract a^3+b^3. The pure cube terms cancel, leaving 3a^2b+3ab^2. Both remaining terms contain 3ab, so factor it out: 3a^2b+3ab^2=3ab(a+b). Thus option A is correct. Option B has an incorrect minus sign; expanding 3ab(a-b) gives 3a^2b-3ab^2, whereas both remaining terms here are positive. Option C omits the coefficient and variables produced by the expansion. Option D is also wrong because only the pure cubes cancel; the two middle terms remain.

In \(m^2-14m+49\), the first and last terms are \(m^2\) and \(7^2\), while the middle term is \(-2\times m\times7=-14m\). Hence it is \((m-7)^2\). Exam tip: check both the middle-term sign and twice-product condition.

Using the identity \((x+y)^2=x^2+y^2+2xy\), we get \(x^2+y^2=(x+y)^2-2xy\). Therefore, \(x^2+y^2=13^2-2(40)=169-80=89\). Option 169 is only \((x+y)^2\); subtracting \(2xy\) is necessary. Exam tip: when the sum and product are given, start with \((x+y)^2-2xy\).

\(x^3+8=x^3+2^3\) is a sum of cubes. Using \(a^3+b^3=(a+b)(a^2-ab+b^2)\), one factor is \(x+2\). \(x-2\) is associated with a difference of cubes. Exam tip: identify the sign before selecting the identity.

In \((a+b)(a-b)\), the terms are the same but the signs of the second terms are opposite. Hence the middle terms cancel, giving \(a^2-b^2\). Exam tip: identify such conjugate binomials first.

Squaring a sum of three terms gives the square of each term and twice every pairwise product. Hence \(2ab\), \(2bc\), and \(2ca\) must all appear. Exam tip: check that no cross-product is missing its factor of 2.

The difference-of-squares identity is \(p^2-q^2=(p+q)(p-q)\). Option B incorrectly treats a sum of squares as a difference. Exam tip: whenever you see \(a^2-b^2\), look for the factors \((a+b)\) and \((a-b)\).

\((a-b)^3=a^3-3a^2b+3ab^2-b^3\), so \(3ab^2\) does appear with a positive sign. Hence none of the listed terms is absent; the options are flawed. Exam tip: for a cube of a difference, signs follow +, −, +, −.

On multiplying \((2x+1)(2x+5)\), we get \(2x\cdot2x=4x^2\), the cross terms \(2x\cdot5+1\cdot2x=12x\), and \(1\cdot5=5\). Therefore, the result is \(4x^2+12x+5\). Option B has an incorrect sum of the middle terms. Exam tip: write all four products while multiplying two binomials, then combine like terms.

Expand two factors first and then multiply by the third. Exam tip: add like terms systematically.

In \((a+b-c)^2\), first use \((a+b)^2=a^2+2ab+b^2\). Adding \(-2c(a+b)+c^2\) gives option A. In option B, the sign of \(2bc\) is incorrect. Exam tip: mixed terms involving the negative term carry a negative sign.

Here \(9p^2=(3p)^2\) and \(4q^2=(2q)^2\). The middle term is \(-12pq=-2(3p)(2q)\), so it matches \((3p-2q)^2\). Option C also forms a square, but it has a positive middle term. Exam tip: take square roots of the end terms, then check the sign of the middle term.

Option A is the expansion of \((a+b+c)^2\): it contains \(a^2,b^2,c^2\) and the pairwise terms \(2ab,2bc,2ca\). In the other options, one mixed term lacks coefficient 2. Exam tip: in a trinomial square, every pairwise product has coefficient 2.