Algebraic identities

The (xy), (yz), and (xz) terms are (6xy), (24yz), and (8xz). Exam tip: find each pair separately.

The number 998 is very close to the convenient base 1000. It can be written as \\(998=1000-2\\), so its square can be found quickly with the identity \\((x-y)^2=x^2-2xy+y^2\\). Substituting \\(x=1000\\) and \\(y=2\\) gives \\(998^2=(1000-2)^2=1000000-4000+4=996004\\). This avoids long multiplication and uses a nearby round number.

Therefore, option B is the most suitable form. Option A, \\((998+2)^2\\), changes the number to 1000 and is not equal to \\(998^2\\). Option C represents 1002 rather than 998, and option D is the difference-of-squares product, not the square of 998. The key is to choose a close base and preserve the subtraction correctly.

Since 998 is 2 less than 1000,
\((998)^2=(1000-2)^2=1000^2-2\times1000\times2+2^2=1000000-4000+4=996004\). Therefore, option A is correct. In option D, the sign of the middle term has effectively been taken incorrectly. Exam tip: use \((a-b)^2\) to square numbers close to 1000.

By difference of squares, ( (1006+994)(1006-994)=2000\cdot12=24000 ). Exam tip: multiply the sum and difference.

The difference-of-squares identity is \(a^2-b^2=(a-b)(a+b)\). On multiplying, the middle terms \(-ab\) and \(+ab\) cancel. In contrast, \((a-b)^2\) contains a \(-2ab\) term. Exam tip: for two squared terms with subtraction, look for the sum-and-difference factors.

This expression has the form \\((a+b)^2-(a-b)^2\\), whose simplification is \\(4ab\\). Set \\(a=8x\\) and \\(b=1\\). Then \\(4ab=4(8x)(1)=32x\\), so option B is correct. The important point is that a represents the complete term \\(8x\\), not just the numerical coefficient 8. This is why the answer is linear in x rather than a constant or a squared expression.

Expanding gives the same confirmation: \\((8x+1)^2=64x^2+16x+1\\), while \\((8x-1)^2=64x^2-16x+1\\). On subtraction, the two \\(64x^2\\) terms cancel, as do the constant 1 terms. The remaining part is \\(16x-(-16x)=32x\\). Option A is only half of the correct difference; options C and D retain terms that should cancel. Hence the second choice follows exactly from the identity.

\(9a^2-16b^2=(3a)^2-(4b)^2\), so it is a difference of squares and factors as \((3a-4b)(3a+4b)\). Option C is a perfect square. Exam tip: first identify the two squared terms.

\(49p^2=(7p)^2\) and \(81q^2=(9q)^2\). Hence, \(49p^2-81q^2=(7p+9q)(7p-9q)\). Option B is a sum, not a difference. In exams, check for two square terms with a minus sign between them.

To find the coefficient of the term containing \\(a^2b\\), use the cube identity \\( (u-v)^3=u^3-3u^2v+3uv^2-v^3\\). Here, \\(u=4a\\) and \\(v=b\\). The required term is the second term, because it contains two factors of \\(u\\) and one factor of \\(v\\). Its value is \\(-3(4a)^2b\\). Since \\((4a)^2=16a^2\\), the term becomes \\(-48a^2b\\).

The coefficient is therefore \\(-48\\), so option A is correct. The negative sign is important because this is a cube of a difference, not a cube of a sum. Option B has the correct magnitude but the wrong sign. Option C results from an incorrect calculation, and option D is the coefficient of the first term \\( (4a)^3\\), not the requested term. Hence both the identity and direct expansion confirm A.

The standard identity is \(a^3+b^3=(a+b)(a^2-ab+b^2)\), with a negative middle term. On multiplying, the cross terms cancel and give \(a^3+b^3\). Exam tip: distinguish the signs in sum and difference of cubes.

The difference of two cubes is \(x^3-y^3=(x-y)(x^2+xy+y^2)\). On multiplying, the middle terms cancel, leaving \(x^3-y^3\). The factor \(x^2-xy+y^2\) is used for the sum of cubes. Exam tip: identify the sign in the outer binomial first.

Recognize the expression as a difference of cubes: \\(a^3-b^3=(a-b)(a^2+ab+b^2)\\). Since 27p³=(3p)³ and 64q³=(4q)³, take a=3p and b=4q. Substitution gives \\( (3p-4q)((3p)^2+(3p)(4q)+(4q)^2)\\), which simplifies to \\( (3p-4q)(9p^2+12pq+16q^2)\\). Hence option D is correct.

The signs provide an important check. For a difference of cubes, the first factor has a minus sign, but every term in the second factor is positive. Multiplying the first terms gives 27p³, and the final product gives 64q³ with subtraction overall. Option A uses the pattern for a sum of cubes, option B uses incorrect cube roots, and option C treats the difference as one cube. The identity confirms the stated factorisation exactly.

The identity for a sum of cubes is \(a^3+b^3=(a+b)(a^2-ab+b^2)\), so \(a+b\) is a factor. The factor \(a-b\) belongs to the difference-of-cubes identity. Exam tip: check the sign between the cubes first.

Expanding \((p-q)^2\) gives \(p^2-2pq+q^2\), so A is correct. In \((p+q)^2\), the middle term is positive. Exam tip: use the sign of the middle term to identify the square.

The identity is \((a+b)^3=a^3+3a^2b+3ab^2+b^3\). Every term has total degree 3, whereas \(a^2b^2\) has total degree 4, so it cannot occur. Exam tip: remember the coefficients 1, 3, 3, 1 for a cube expansion.

Use the identity \((m+n)^2=m^2+n^2+2mn\). Substituting the given values gives \(14^2=100+2mn\), so \(196=100+2mn\). Hence, \(2mn=96\) and \(mn=48\). Option 96 is a close distractor because it is the value of \(2mn\), not \(mn\). Exam tip: after finding \(2mn\) from this identity, divide by 2 to obtain \(mn\).

The identity is \((x-y)^2=x^2-2xy+y^2\), so option B is correct. Option C comes from \((x+y)(x-y)\), not a square. Exam tip: check the sign of the middle term carefully.

The identity is \((a-b)^3=a^3-3a^2b+3ab^2-b^3\). The middle terms have negative and then positive signs; option C incorrectly makes \(3ab^2\) negative. Exam tip: remember the sign pattern \(+,-,+,-\) for a difference cube.

The expression is a product of three linear factors. Its complete expansion must contain a cubic term, a squared term, a linear term, and a constant term. The safest method is to multiply two factors first and then use distribution carefully. This also helps prevent missing a middle term or placing a coefficient with the wrong power of x.

First, \\(x+3)(x+5)=x^2+8x+15\\). Now multiply by \\(x+7\\): \\(x^2(x+7)+8x(x+7)+15(x+7)\\), which gives \\(x^3+7x^2+8x^2+56x+15x+105\\). Combining like terms gives \\(x^3+15x^2+71x+105\\). Therefore option A is correct. Option B misses part of the squared coefficient, while C and D omit or interchange terms.

First multiply the two outer binomials: \((2x+1)(2x+9)=4x^2+18x+2x+9=4x^2+20x+9\). Therefore, option A is correct. In option B, the middle terms from \(2x\cdot9\) and \(1\cdot2x\) have not been added correctly. Exam tip: write all four products before combining like terms.