Algebraic identities

First, \((x+1)(x+2)=x^2+3x+2\). Multiplying this by \((x+3)\) gives \(x^3+3x^2+3x^2+9x+2x+6=x^3+6x^2+11x+6\). Hence, option A is correct. In option B, the coefficient of \(x^2\) is incorrectly written as 5 instead of 6. Exam tip: combine like terms carefully after multiplication.

The direct answer is option B: \\(a^3+3a^2b+3ab^2+b^3\\). The cube means three factors: \\(a+b\\)^3=(a+b)(a+b)(a+b)\\). The binomial cube identity is \\(x+y\\)^3=x^3+3x^2y+3xy^2+y^3\\). Substituting \\(x=a\\) and \\(y=b\\) gives \\(a^3+3a^2b+3ab^2+b^3\\). Option B is correct because it has the two middle terms, both with coefficient 3, and all signs are positive. Option A is wrong because it omits the powers and does not contain the complete middle terms. Option C is the identity for \\(a-b\\)^3\\), so its alternating signs do not fit a sum. Option D contains only the end cubes and leaves out the middle terms. A useful check is to put \\(a=b=1\\): the left side is \\(2^3=8\\), and option B gives \\(1+3+3+1=8\\). Remember: plus-cube has four positive terms with pattern 1, 3, 3, 1.

Expanding \((p+q)^2\) gives \(p^2+2pq+q^2\), so A is the square of a binomial sum. B represents \((p-q)^2\). Exam tip: check that the middle term is exactly \(2pq\).

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.

The direct answer is A: \(64m^3-125n^3=(4m-5n)(16m^2+20mn+25n^2)\). First recognise \(64m^3=(4m)^3\) and \(125n^3=(5n)^3\). Use the difference-of-cubes identity \(a^3-b^3=(a-b)(a^2+ab+b^2)\). Put \(a=4m\) and \(b=5n\). Then the factorisation is \((4m-5n)((4m)^2+(4m)(5n)+(5n)^2)\), which becomes \((4m-5n)(16m^2+20mn+25n^2)\). Option A is exactly this result. Option B uses the sum-of-cubes pattern and would represent \((4m)^3+(5n)^3\), not a difference. Option C has incorrect cube roots: \(8m\) cubed is \(512m^3\), not \(64m^3\), and its second factor is also wrong. Option D means \((4m-5n)^3\), which creates extra terms and is not the same as a simple difference of cubes. Memory cue: for \(a^3-b^3\), use “minus outside, all plus inside.”

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 direct answer is option A: \(30x^2+250\). Expand both cubes carefully. Using \((x+5)^3=x^3+15x^2+75x+125\), and \((x-5)^3=x^3-15x^2+75x-125\), subtract the second expression from the first: \(x^3-x^3=0\), \(15x^2-(-15x^2)=30x^2\), \(75x-75x=0\), and \(125-(-125)=250\). Therefore the result is \(30x^2+250\). Option A is correct because it contains the surviving square and constant terms. Option B, \(30x^2+50x+250\), is wrong because the linear terms are actually equal and cancel; no \(x\) term remains. Option C incorrectly leaves a cubic term and uses an incorrect coefficient. Option D gives only the constant part and omits \(30x^2\). A useful check is that replacing \(x o -x\) changes the original expression to its negative, so the result should contain only odd-power behaviour overall; because the difference is arranged with the shifts, the calculated expression confirms the expansion directly. Exam cue: expand signs carefully, especially when subtracting the entire second bracket.

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.

Using ( (x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx) ), (36=x^2+y^2+z^2+22). Exam tip: apply the three-term identity directly.

Use the sum of cubes identity \(a^3+b^3=(a+b)(a^2-ab+b^2)\). Here \(8x^3=(2x)^3\) and \(27=3^3\). So
\(8x^3+27=(2x)^3+3^3=(2x+3)((2x)^2-(2x)(3)+3^2)\)
\(=(2x+3)(4x^2-6x+9)\).

Using \((a+b)^2+(a-b)^2=2a^2+2b^2\), take \(a=2x\) and \(b=3y\). This gives \(2(2x)^2+2(3y)^2=8x^2+18y^2\). The \(24xy\) terms cancel when the two squares are added, so option C is not correct. Exam tip: recognise the \((a+b)^2+(a-b)^2\) pattern and apply the identity directly.