Algebraic identities

The direct answer is B, 24x^2. Use the cube identities or expand carefully. The x-squared terms come from the terms 3x^2(4) in \\((x+4)^3\\) and 3x^2(-4) in \\((x-4)^3\\). Since the second cube is subtracted, its x-squared contribution is also subtracted: \\(12x^2-(-12x^2)=24x^2\\). Equivalently, \\((x+a)^3-(x-a)^3=6ax^2+2a^3\\), and a=4 gives 24x^2+128. Option A, 12x^2, counts only one contribution. Option B is correct because both opposite middle contributions combine. Option C, 48x^2, doubles the correct result, and option D, 96x^2, is an even larger coefficient error. Only the x-squared term is requested, not the complete expression. Exam cue: in a cube difference with ±a, the x^2 coefficient is 6a.

In the sum, constant cube terms cancel and (2x^3+6x\cdot4=2x^3+24x) remains. Exam tip: understand symmetric cube expansion.

Using \((a+b)^2=a^2+2ab+b^2\), subtracting \(a^2-b^2\) changes the signs of both terms inside it. Thus, \(a^2+2ab+b^2-a^2+b^2=2ab+2b^2\). Option B misses the remaining \(2b^2\) term. Exam tip: when a bracket is preceded by a minus sign, change the sign of every term in that bracket.

( (a-b)^2=a^2-2ab+b^2 ), so subtracting leaves (-2ab). Exam tip: remember the negative middle term.

\(x^2+2xy+y^2=(x+y)^2\), a perfect square. The square of every real number is never negative, so this expression is always non-negative. Exam tip: recognise the middle term \(2xy\).

On squaring, (t^2-2+\frac{1}{t^2}=36), so the value is (38). Exam tip: the square of a difference gives (-2).

This is (4AB) where (A=x) and (B=\frac{1}{x}), so the value is (4). Exam tip: reciprocal terms multiply to (1).

The direct answer is option C: \(x^4-16\). Start with the identity \((a-b)(a+b)=a^2-b^2\). Here, \((x-2)(x+2)=x^2-4\). Now apply the same identity again: \((x^2-4)(x^2+4)=(x^2)^2-4^2=x^4-16\). Therefore C is correct. Option A, \(x^4+16\), incorrectly changes subtraction into addition. Option B, \(x^4-8\), uses the wrong constant because \(4^2=16\), not 8. Option C follows both difference-of-squares steps exactly. Option D, \(x^4+8x^2+16\), is the expansion of \((x^2+4)^2\), not the given product. Memory cue: opposite signs in both pairs mean difference of squares; square the two terms and subtract.

In \((a+b)(a-b)\), the cross terms \(+ab\) and \(-ab\) cancel, leaving \(a^2-b^2\). Option A is the square of a sum, not this product. Exam tip: look for opposite signs.

Since \(p^4=(p^2)^2\) and \(81q^4=(9q^2)^2\), option A is a difference of two squares. It factorises as \((p^2-9q^2)(p^2+9q^2)\). Exam tip: first rewrite each term as a perfect square.

This is a difference of two perfect squares: \(49p^2=(7p)^2\) and \(64q^2=(8q)^2\). Hence the identity \(a^2-b^2\) applies. Exam tip: first check whether both terms are perfect squares.

The correct identity is \(p^2-q^2=(p+q)(p-q)\). On multiplying, the middle terms \(-pq\) and \(+pq\) cancel, leaving \(p^2-q^2\). Exam tip: check cancellation of middle terms.

\(9x^2-12xy+4y^2=(3x-2y)^2\). Its middle term is \(-2(3x)(2y)=-12xy\). In option B, the last term is not \(4y^2\). Exam tip: check the middle term using twice the product of the square roots.

In the square of three terms, the mixed term formed by two selected terms is twice their product. Here the relevant terms are 2x and z, so the xz term is

2(2x)(z)=4xz.

The term -6yz comes from the product of -3y and z, so it is not an xz term. Exam tip: identify the required pair of terms and take twice their product.

In \(x^2+6x+9\), the middle term is \(2\cdot x\cdot3=6x\) and the last term is \(3^2=9\), so it equals \((x+3)^2\). \(x^2+6x+8\) fails because its constant term is not 9. Exam tip: check \(a^2+2ab+b^2\).

The constant term (-4k=-28) gives (k=7), and the middle term also becomes (3x). Exam tip: verify using both conditions.

The first is (x^2+6x+9) and the second is (x^2+6x-7). Subtracting gives (9-(-7)=16), so the correct option should be (16).

The first product is (4x^2+20x+9) and the second is (4x^2+20x+25). Exam tip: cancel like terms and compare constants.

The correct identity is \(a^2-b^2=(a+b)(a-b)\). Multiplying gives \(a^2-ab+ab-b^2=a^2-b^2\). In contrast, \((a-b)^2\) contains an extra \(-2ab\) term. Exam tip: look for conjugate binomials with opposite signs.

The direct answer is C: \\((5a-4b)^2+(5a+4b)^2=50a^2+32b^2\\). Use \\((p-q)^2=p^2-2pq+q^2\\) and \\((p+q)^2=p^2+2pq+q^2\\), with \\(p=5a\\) and \\(q=4b\\). The first square is \\(25a^2-40ab+16b^2\\); the second is \\(25a^2+40ab+16b^2\\). Adding gives \\(50a^2+32b^2\\), because \\(-40ab+40ab=0\\). Therefore C is correct. Option A, \\(25a^2+16b^2\\), is only one square’s non-middle terms, so it misses the doubling. Option B, \\(40ab\\), is a middle term that cancels. Option D has the wrong negative sign for the b-squared term; squared terms add positively. Memory cue: adding opposite squares cancels the middle terms.

The expression contains two squares that differ only in the sign of the middle term. Group the terms that are unchanged by writing the expression as \(A+y)^2-(A-y)^2\), where \(A=x+z\). The identity \( (A+B)^2-(A-B)^2=4AB\) says that this difference is four times the product of the common part and the changing term. This avoids expanding every term separately.

Substituting \(A=x+z\) and \(B=y\) gives \(4(x+z)y=4y(x+z)\). Thus the correct simplified form is option A. Option B is exactly half the required value, option C keeps only an incomplete product, and option D does not represent the difference of these two squares. The grouping and identity also provide a quick check on the result.

A three-term square has double products of every pair. Here the (pq), (pr), and (qr) terms are (12pq), (6pr), and (4qr).

Let the perfect square be \((ax-by)^2\). Then
\[
(ax-by)^2=a^2x^2-2abxy+b^2y^2.
\]
Match with \(16x^2-40xy+25y^2\):
\[
a^2=16\Rightarrow a=4,
\]
\[
b^2=25\Rightarrow b=5.
\]
Check the middle term:
\[
-2ab=-2\cdot 4\cdot 5=-40,
\]
which matches \(-40xy\). Hence the expression equals \((4x-5y)^2\).

\(81a^2-49b^2=(9a)^2-(7b)^2\). Applying the identity \(x^2-y^2=(x+y)(x-y)\) gives \((9a+7b)(9a-7b)\). Options B and D are perfect squares and would produce a middle term of \(\mp126ab\), which is absent in the given expression. Exam tip: identify the square roots of both terms, then write one sum factor and one difference factor.

The sum of constants is (12+(-5)=7) and the product is (-60). Exam tip: check sum and product separately with mixed signs.