Algebraic identities

Using \((a+b)^3=a^3+3a^2b+3ab^2+b^3\), put \(a=x\) and \(b=2\). Thus, \((x+2)^3=x^3+3x^2(2)+3x(2^2)+2^3=x^3+6x^2+12x+8\). In option B, the coefficients of the middle terms are incorrect, while option C omits both middle terms. Exam tip: in a cube expansion, always include both middle terms, \(3a^2b\) and \(3ab^2\).

In the cube of a difference, signs occur in (+), (-), (+), (-) order. Exam tip: write the cube of (2x) as (8x^3).

The identity for the sum of cubes is \(a^3+b^3=(a+b)(a^2-ab+b^2)\), so option A is correct. \((a-b)(a^2+ab+b^2)\) is the factorisation of the difference of cubes, \(a^3-b^3\), so option B is not correct. Exam tip: for a sum of cubes, use \(+\) in the first bracket and \(-\) for the middle term in the second bracket.

The difference of cubes identity is \\(a^3-b^3=(a-b)(a^2+ab+b^2)\\). Applying it directly gives \\(a^3-b^3=(a-b)(a^2+ab+b^2)\\). Multiplying the factors confirms the result: the terms combine to \\(a^3-ab^2+a^2b+ab^2-b^3=a^3-b^3\\).

Therefore, option A is correct. The factor outside the bracket has a minus sign, while all three terms inside the second factor have positive signs. Option B is the identity for a sum of cubes, not a difference. Option C is the cube of a difference and contains different middle terms, while option D concerns squares rather than cubes. Keeping the degree and signs distinct avoids confusion.

Since \(27x^3+8y^3=(3x)^3+(2y)^3\), use \(a^3+b^3=(a+b)(a^2-ab+b^2)\). Substituting \(a=3x\) and \(b=2y\) gives \((3x+2y)(9x^2-6xy+4y^2)\). Option B uses the signs for the difference-of-cubes identity, so it represents \(27x^3-8y^3\), not the given expression. Exam tip: for a sum of cubes, the middle term in the second factor is negative.

This is ( (4m)^3-(5n)^3 ). Exam tip: in difference of cubes, the second factor has all positive terms.

Use the identity \((a+b)^3=a^3+3a^2b+3ab^2+b^3\). After subtracting \(a^3+b^3\), the remaining expression is \(3a^2b+3ab^2\). Taking \(3ab\) as the common factor gives \(3ab(a+b)\). In \(3ab(a-b)\), the second middle term would be negative, so it is incorrect. Exam tip: while expanding a cube, check the signs of both middle terms carefully.

The sum of two cubes factorises as \((a+b)(a^2-ab+b^2)\). On multiplying, the middle \(ab\) terms cancel, leaving \(a^3+b^3\). Exam tip: for a sum, use \(a+b\) first and a negative middle term in the quadratic factor.

The correct identity is \((a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca\). On squaring, every distinct pair occurs twice. Option A misses the factor 2. Exam tip: check all three pairs: ab, bc, and ca.

Using the identity \((a-b)^2=a^2-2ab+b^2\), we get \(a^2+b^2=(a-b)^2+2ab=6^2+2\times16=36+32=68\). Option 36 is only \((a-b)^2\); the term \(2ab\) must also be added. Exam tip: for \(a^2+b^2\), add \(2ab\) to \((a-b)^2\).

The square contains the square of each term and twice the product of every pair of terms. Hence \(2xy,2yz,2zx\) must appear. Exam tip: list all three pairs before choosing an identity.

Use the identity \((m+n)^2-(m-n)^2=4mn\). Substituting the given values gives \(12^2-4^2=4mn\), i.e., \(144-16=4mn\). Hence \(128=4mn\), so \(mn=32\). Option 16 may result from an incorrect division of 128 by 4. Exam tip: In such questions, square the two given expressions first and then take their difference.

On expanding cubes, (+15x^2) and (-(-15x^2)) combine to give (30x^2). Exam tip: signs change in subtraction.

The (x^3) and (x) terms of both cubes cancel. Exam tip: expand fully and cancel like terms.

Use \(a^3+b^3=(a+b)(a^2-ab+b^2)\). Here, \(a=x\) and \(b=2y\), so \(x+2y\) is a factor. \(x-2y\) is associated with a difference of cubes. Exam tip: identify the cube bases first.

Use the identity


a+b)^3=a^3+3a^2b+3ab^2+b^3
a-b)^3=a^3-3a^2b+3ab^2-b^3
with the first term equal to 2a. Thus,


(2a+b)^3=(2a)^3+3(2a)^2b+3(2a)b^2+b^3
=8a^3+12a^2b+6ab^2+b^3.
Therefore, the coefficient of a^2b is 12. The value 6 is the coefficient of ab^2, so it is a close but incorrect distractor. Exam tip: while finding a coefficient, include the numerical factor raised to the required power.

Expand the square of three terms using \\(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca\\). Substituting the given terms, the first expression becomes \\(a^2+b^2+c^2+2ab+2bc+2ca\\). Now subtract \\(a^2+b^2+c^2+2ab\\). These four terms cancel exactly, leaving only the two cross-products \\(2bc+2ca\\).

Thus option A is correct. The term \\(2ab\\) is removed because it appears in the quantity being subtracted, and \\(abc\\) cannot appear because squaring a sum produces squared terms and pairwise products, not a product of all three variables. The result may also be written as \\(2c(a+b)\\), which is algebraically equivalent to \\(2bc+2ca\\).

The (xz) term is (2\cdot x\cdot3z=6xz) and the (yz) term is (2\cdot2y\cdot3z=12yz). Exam tip: find each pair separately.

( (a+b)^2=a^2+2ab+b^2 ), so equality needs (2ab=0). Exam tip: understand the role of the middle term.

Use the identity \((a+b)^2-(a-b)^2=4ab\). Here, \(a=2x\) and \(b=y\), so the expression becomes \(4\times 2x\times y=8xy\). \(4xy\) is a close distractor, but it misses the factor 2 in \(a=2x\). Exam tip: recognise the \(4ab\) identity directly in such expressions.