Algebraic identities

\((p+q)(p-q)\) are conjugate binomials, and their product is \(p^2-q^2\), a difference of two squares. In \((p+q)^2\), the middle term is \(2pq\). Exam tip: look for the same terms with opposite signs.

Use the identity
\((a-b)^2=a^2-2ab+b^2\). Thus,
\((4m-3n)^2=16m^2-24mn+9n^2\). Subtracting
\((16m^2+9n^2)\) cancels the square terms, leaving
\(-24mn\).
\(24mn\) has the wrong sign because the middle term in a square of a difference is negative. Exam tip: in
\((a-b)^2\), the middle term is always
\(-2ab\).

Use the identities \((a+b)^2=a^2+2ab+b^2\) and \((a-b)^2=a^2-2ab+b^2\). Their sum is \(2a^2+2b^2\), since the \(+2ab\) and \(-2ab\) terms cancel. Subtracting \(2a^2+2b^2\) therefore gives \(0\). \(2ab\) is not correct because the middle terms add to zero. Exam tip: remember \((x+y)^2+(x-y)^2=2x^2+2y^2\).

Both expansions have the same (x^2+14x), and constants give (40-48=-8). Exam tip: cancel like terms and subtract constants.

The pair terms in the square are (6ab), (6ac), and (2bc). Exam tip: take the double product of every pair.

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.

Use \(a^3-b^3=(a-b)(a^2+ab+b^2)\). Here \(27p^3=(3p)^3\) and \(64q^3=(4q)^3\). So
\(27p^3-64q^3=(3p)^3-(4q)^3=(3p-4q)\big((3p)^2+(3p)(4q)+(4q)^2\big)\)
\(=(3p-4q)(9p^2+12pq+16q^2)\).

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.