Algebraic identities

\(x^3+8=x^3+2^3\) is a sum of cubes. Using \(a^3+b^3=(a+b)(a^2-ab+b^2)\), one factor is \(x+2\). \(x-2\) is associated with a difference of cubes. Exam tip: identify the sign before selecting the identity.

In \((a+b)(a-b)\), the terms are the same but the signs of the second terms are opposite. Hence the middle terms cancel, giving \(a^2-b^2\). Exam tip: identify such conjugate binomials first.

Squaring a sum of three terms gives the square of each term and twice every pairwise product. Hence \(2ab\), \(2bc\), and \(2ca\) must all appear. Exam tip: check that no cross-product is missing its factor of 2.

The difference-of-squares identity is \(p^2-q^2=(p+q)(p-q)\). Option B incorrectly treats a sum of squares as a difference. Exam tip: whenever you see \(a^2-b^2\), look for the factors \((a+b)\) and \((a-b)\).

\((a-b)^3=a^3-3a^2b+3ab^2-b^3\), so \(3ab^2\) does appear with a positive sign. Hence none of the listed terms is absent; the options are flawed. Exam tip: for a cube of a difference, signs follow +, −, +, −.

On multiplying \((2x+1)(2x+5)\), we get \(2x\cdot2x=4x^2\), the cross terms \(2x\cdot5+1\cdot2x=12x\), and \(1\cdot5=5\). Therefore, the result is \(4x^2+12x+5\). Option B has an incorrect sum of the middle terms. Exam tip: write all four products while multiplying two binomials, then combine like terms.

The direct answer is option A: \(x^3+12x^2+44x+48\). Start by multiplying the first two factors: \((x+2)(x+4)=x^2+6x+8\). Now multiply by \((x+6)\): \(x^2(x+6)+6x(x+6)+8(x+6)=x^3+6x^2+6x^2+36x+8x+48\). Combining like terms gives \(x^3+12x^2+44x+48\). Option A is correct because it includes every product and combines the terms correctly. Option B has \(10x^2\), but the two contributions to the squared term are \(6x^2+6x^2=12x^2\). Option C misses the \(x^2\) term and has an incorrect linear term. Option D places the coefficients of \(x^2\) and \(x\) incorrectly. Multiply in two stages and collect like powers carefully. Memory cue: the constant term is \(2\cdot4\cdot6=48\), and the coefficient of \(x^2\) is \(2+4+6=12\).

In \((a+b-c)^2\), first use \((a+b)^2=a^2+2ab+b^2\). Adding \(-2c(a+b)+c^2\) gives option A. In option B, the sign of \(2bc\) is incorrect. Exam tip: mixed terms involving the negative term carry a negative sign.

Here \(9p^2=(3p)^2\) and \(4q^2=(2q)^2\). The middle term is \(-12pq=-2(3p)(2q)\), so it matches \((3p-2q)^2\). Option C also forms a square, but it has a positive middle term. Exam tip: take square roots of the end terms, then check the sign of the middle term.

Option A is the expansion of \((a+b+c)^2\): it contains \(a^2,b^2,c^2\) and the pairwise terms \(2ab,2bc,2ca\). In the other options, one mixed term lacks coefficient 2. Exam tip: in a trinomial square, every pairwise product has coefficient 2.

\(x^2+2xy+y^2=(x+y)^2\), so it is a perfect-square trinomial. Check the middle term: \(2xy=2\cdot x\cdot y\). In option B, the middle term is only \(xy\), so it cannot be \((x+y)^2\). Exam tip: compare the middle term with twice the product of the square roots of the end terms.

Since \(81m^4=(9m^2)^2\) and \(16n^6=(4n^3)^2\), option A has the form \(a^2-b^2\) and factors as \((9m^2-4n^3)(9m^2+4n^3)\). Option B is a sum, not a difference. Exam tip: check for two square terms separated by a minus sign.

In the cube of a difference, signs occur in (+), (-), (+), (-) order. Exam tip: check both (3^2) and (3^3).

( (2x)^3=8x^3 ), (3(2x)^2\cdot3=36x^2), and (3(2x)\cdot9=54x). Exam tip: check the power of every term.

\(9x^2-24xy+16y^2=(3x-4y)^2\), so it has two identical binomial factors. Its middle term is \(-2(3x)(4y)=-24xy\). In option B, the last term would need to be \(16y^2\). Exam tip: check square roots of the first and last terms.

The expression is a difference of two cubes because \\(125p^3=(5p)^3\\) and \\(8q^3=(2q)^3\\). The identity for a difference of cubes is \\(a^3-b^3=(a-b)(a^2+ab+b^2)\\). Substituting \\(a=5p\\) and \\(b=2q\\) gives \\((5p-2q)(25p^2+10pq+4q^2)\\).

Thus option A is correct. The second factor contains a plus sign between all three terms; this is a key feature of the difference-of-cubes identity. Option B uses the corresponding sum-of-cubes pattern, while C uses incorrect cube roots and D treats the whole expression as the cube of a binomial. Expansion of option A returns the original expression.

The cube identity is \((a+b)^3=a^3+3a^2b+3ab^2+b^3\), so A is correct. In B, the sign of the last term is wrong. Exam tip: remember the coefficient pattern 1, 3, 3, 1.

The end terms have roots \(2p\) and \(-3q\); twice their product is \(-12pq\). Therefore, A is \((2p-3q)^2\). C fails the middle-term test. Exam tip: always verify the middle coefficient.

\(4x^2-12xy+9y^2=(2x)^2-2(2x)(3y)+(3y)^2=(2x-3y)^2\), so A is a perfect square. In B, the middle term should be \(-12xy\), not \(-10xy\). Exam tip: check the middle term using twice the product of square roots.

(81=35+2(ab+bc+ca)), so (2(ab+bc+ca)=46) and the value is (23). Exam tip: find the double product sum first.

The (xy) term is (12xy), the (yz) term is (6yz), and the (xz) term is (4xz). Exam tip: find each double product separately.

Most errors occur due to incomplete terms, wrong signs, or wrong powers. Exam tip: identify the whole term before applying the formula.

\((u-v)^2=u^2-2uv+v^2\), so this trinomial is the square of a difference. In contrast, \((u+v)^2\) has the middle term \(+2uv\). Exam tip: always check the sign of the middle term.

First expand \((x+2y+z)^2\):
\[(x+2y+z)^2=x^2+(2y)^2+z^2+2\cdot x\cdot 2y+2\cdot x\cdot z+2\cdot 2y\cdot z\]
\[=x^2+4y^2+z^2+4xy+2xz+4yz\]
Now subtract \(x^2+4y^2+z^2\):
\[(x^2+4y^2+z^2+4xy+2xz+4yz)-(x^2+4y^2+z^2)=4xy+2xz+4yz\]
So the simplified form is \(4xy+4yz+2xz\) (same terms).