Algebraic identities

Using \( (a+b)^2=a^2+2ab+b^2 \), with \(a=x\) and \(b=3y\), we get \( (x+3y)^2=x^2+6xy+9y^2 \). Thus, the \(x^2\), \(6xy\), and \(9y^2\) terms all cancel in the given expression, leaving \(0\). \(6xy\) is only the middle term in the expansion, not the final value. Exam tip: always check the middle term \(2ab\) when expanding a square.

The identity \((a+b)^2=a^2+2ab+b^2\) applies here. Taking \(a=x\) and \(b=5y\), the middle term is \(2\times x\times5y=10xy\). Option C represents a square of a difference. Exam tip: always check both the sign and coefficient of the middle term.

In \((a-b)^3=a^3-3a^2b+3ab^2-b^3\), the terms \(-3a^2b\) and \(-b^3\) contain odd powers of \(b\), so their coefficients are negative. The \(b^2\) term is positive. Exam tip: track signs by parity.

In the cube of a difference, signs follow (+), (-), (+), (-). Exam tip: check both (4^2) and (4^3).

( (3x)^3=27x^3 ), (3(3x)^2\cdot4=108x^2), and (3(3x)\cdot16=144x). Exam tip: check the power of every term.

\(x^2-49=x^2-7^2\), so it factors as \((x+7)(x-7)\). The factors differ only by sign. \(x^2+49\) is a sum of squares. Exam tip: first check whether both terms are perfect squares.

This is again a difference of cubes. The first term is 216p³=(6p)³, and the second is 125q³=(5q)³. Applying \\(a^3-b^3=(a-b)(a^2+ab+b^2)\\) with a=6p and b=5q gives \\( (6p-5q)(36p^2+30pq+25q^2)\\). Therefore option A is correct.

To verify, the first factor supplies the difference of the cube roots, while the second factor contains the squares and positive cross-product: \\( (6p)^2=36p^2\\), \\( (6p)(5q)=30pq\\), and \\( (5q)^2=25q^2\\). The second factor must have all positive signs for a difference of cubes. Option B has the wrong signs, option C uses incorrect cube roots, and option D is the cube of a binomial rather than the difference-of-cubes identity. The supplied factorisation matches the standard rule exactly.

\(x^3-27=x^3-3^3\), so it is a difference of two perfect cubes. Using \(a^3-b^3=(a-b)(a^2+ab+b^2)\), it factorises as \((x-3)(x^2+3x+9)\). In contrast, \(x^3+27\) is a sum of cubes. Exam tip: recognise \(27=3^3\).

The difference-of-cubes identity is \(a^3-b^3=(a-b)(a^2+ab+b^2)\). For \(a^3+b^3\), the first factor is \(a+b\). Exam tip: in the difference of cubes, the middle term of the quadratic factor is positive.

\((x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2zx\). It contains square terms and products of pairs of variables only; \(xyz\) is a product of three variables, so it is absent. Exam tip: list every pairwise product with coefficient \(2\).

\(4a^2-12ab+9b^2=(2a-3b)^2\). Its middle term is \(-2\times2a\times3b=-12ab\). Option D has \(-6ab\), so it is not a perfect square. Exam tip: compare the middle term with twice the product of the square roots of the outer terms.

Option A contains conjugate binomials with the same terms and opposite signs. Using \((A+B)(A-B)=A^2-B^2\), it becomes \((7m)^2-(3n)^2\). The other options are perfect squares. Exam tip: look for opposite signs in conjugate factors.

The difference of the first two squares is (24x), and ( (x+5)(x-5)=x^2-25 ). Exam tip: combine identities while solving.

The required term is the part containing exactly one a and two b factors. In a cube such as \\( (u+v)^3\\), the mixed term containing \\(uv^2\\) is \\(3uv^2\\). Here the two cubes have opposite signs in their middle terms, and the entire second cube is subtracted. Consequently, the two contributions add rather than cancel.

For the first cube, take \\(u=4a\\) and \\(v=b\\). Its required term is \\(3(4a)b^2=12ab^2\\). In \\( (4a-b)^3\\), the corresponding term is \\(-12ab^2\\). Since the expression subtracts this whole cube, the contribution becomes \\(-(-12ab^2)=+12ab^2\\). Thus the total is \\(12ab^2+12ab^2=24ab^2\\). Option A is correct. Option B counts only one cube and therefore gives half the result.

Most mistakes happen due to incomplete terms or wrong signs. Exam tip: identify the terms before applying the formula.

The first expansion is (x^2+12x+36) and the second is (x^2+12x+20). Exam tip: cancel like terms and compare constants.

Use \(a^3+b^3=(a+b)^3-3ab(a+b)\) where \(ab=1\). Exam tip: remember \(7^3-3\times7=322\).

This is ((3a)^3+(2b)^3), so the sum of cubes identity applies. Exam tip: the middle term inside the second factor is negative.

(a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=81-40=41). Exam tip: directly use the square identity of three terms.

Use ( (a+b)^3+(a-b)^3=2a^3+6ab^2 ). Here (a=x) and (b=2y), so the result is (2x^3+24xy^2).