Algebraic identities

The direct answer is A: 50a² + 8b². Use the identity \\(u+v\\)^2+\\(u-v\\)^2=2u^2+2v^2. Here, u = 5a and v = 2b. Therefore the expression becomes 2\\(5a\\)^2 + 2\\(2b\\)^2 = 2(25a²) + 2(4b²) = 50a² + 8b². If expanded separately, the first square has +20ab and the second has -20ab; these middle terms cancel. Option A is correct. Option B, 25a² + 4b², is only half of the required result because both squares contribute. Option C, 20ab, is not the final value; the opposite middle terms cancel instead of adding. Option D, 50a² - 8b², has the wrong sign for the b² term; squared terms contribute positively here. A useful check is to put a = 1 and b = 1: the original value is 7² + 3² = 49 + 9 = 58, and option A also gives 50 + 8 = 58. Memory trick: in the sum of opposite squares, middle terms disappear and the square terms double.

The direct answer is B: the sum is \\(4pq+4pr\\). Use the identity \\((a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca\\). Here the three parts are \\(a=2p\\), \\(b=q\\), and \\(c=r\\). The pair containing p and q gives \\(2(2p)(q)=4pq\\). The pair containing p and r gives \\(2(2p)(r)=4pr\\). Therefore their sum is \\(4pq+4pr\\), which may also be written \\(4p(q+r)\\). Option A is wrong because it misses the factor 2 in each cross-product. Option B is correct because both required pair terms have been doubled correctly. Option C is wrong as the coefficient is only half of the required result. Option D is wrong because \\(pqr\\) is a three-variable product, whereas squaring a trinomial produces pairwise cross-products, not a three-variable product. Memory cue: in a square, every pair gets a double product.

(9x^2=(3x)^2), (25y^2=(5y)^2), and the middle term is (30xy). Exam tip: check both square roots and the middle term.

Using \((A-B)^2=A^2-2AB+B^2\), take \(A=5x\) and \(B=3y\). This gives \(25x^2-30xy+9y^2\). Option C has a positive middle term, so it represents \((A+B)^2\). Exam tip: always check the sign of the middle term.

The sum of constants is (9+(-4)=5) and the product is (-36). Exam tip: check sum and product separately in mixed signs.

Using \((a+b)^2=a^2+2ab+b^2\), \((2x+3)^2=4x^2+12x+9\). Subtracting \(4x^2+9\) cancels the \(4x^2\) and \(9\) terms, leaving \(12x\). \(6x\) is a close distractor because the middle term is \(2\times 2x\times 3=12x\). Exam tip: always calculate the middle term of a binomial square as \(2ab\).

It matches \(a^2-2ab+b^2=(a-b)^2\), with \(a=p\) and \(b=5q\). Check the middle term: \(-2\times p\times5q=-10pq\). Perfect-square outer terms alone are not enough. In exams, always verify the \(\pm2ab\) term.

The direct answer is A: 0. Apply the identity \\(p+q\\)^2-\\(p-q\\)^2=4pq. Thus the first two parts together equal 4pq. The full expression is then 4pq - 4pq = 0. By direct expansion, \\(p+q\\)^2=p^2+2pq+q^2 and \\(p-q\\)^2=p^2-2pq+q^2; subtraction cancels p² and q² and leaves 4pq. Option A is correct because the remaining term is exactly cancelled. Option B, 4pq, is only the value of the first difference before the final subtraction, so it is incomplete. Option C, 2pq, results from missing one of the two middle terms and is therefore an arithmetic error. Option D, p²-q², does not follow from this difference-of-squares expression; those terms cancel. A quick check with p = 2 and q = 1 gives 3² - 1² - 8 = 9 - 1 - 8 = 0. Memory cue: first difference gives 4pq, then the outside minus 4pq removes it completely.

\(p^2+16q^2=p^2+(4q)^2\) is a sum of squares, whereas the identity is \(A^2-B^2=(A+B)(A-B)\). In \(9a^2-b^2\), the terms are subtracted. Exam tip: check the sign between the two square terms first.

The direct answer is C: -3. Expand each product carefully. First, \\(x+3\\)(x+7)=x^2+7x+3x+21=x^2+10x+21. Second, \\(x+4\\)(x+6)=x^2+6x+4x+24=x^2+10x+24. Now subtract the complete second expression: \\(x^2+10x+21\\)-(x^2+10x+24). Like terms cancel: x²-x²=0 and 10x-10x=0. The constants give 21-24=-3. Therefore option C is correct. Option A, 3, has the sign reversed; it would come from subtracting the first constant from the second. Option B, x, is wrong because the x terms are equal and cancel. Option C correctly keeps the order of subtraction and gives -3. Option D, 0, notices cancellation of variable terms but forgets that the constants are different. A useful check is to take x = 0: the expression becomes 3×7 - 4×6 = 21 - 24 = -3, confirming that the result does not depend on x. Memory cue: expand, cancel like terms, then preserve the subtraction order.

The pair terms in the square are (4ab), (2bc), and (4ac). Exam tip: take the double product of each pair.

Using \((a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca\), the term containing \(yz\) comes from \(2(2y)(5z)\). Thus, \(2\times2\times5\,yz=20yz\), so \(20yz\) is correct. \(10yz\) results from forgetting the factor \(2\) in the cross-product term. Exam tip: while squaring three terms, multiply every product of two unlike terms by \(2\).

Since \((a+b)^2=(a+b)(a+b)\), multiplication gives \(a^2+ab+ab+b^2=a^2+2ab+b^2\). Option B is the expansion of \((a-b)^2\). Exam tip: the middle term is always \(2ab\).

Write 997 as \(1000-3\). Using \((a-b)^2=a^2-2ab+b^2\), \((1000-3)^2=1000^2-2\times1000\times3+3^2=1000000-6000+9=994009\). Therefore, the correct answer is 994009. The value 997009 results from an incorrect middle-term calculation. Exam tip: use \((a-b)^2\) for squaring numbers close to 1000.

The direct answer is B, 16000. Use the difference-of-squares identity: \\(a^2-b^2=(a+b)(a-b)\\). Here, take a=1004 and b=996. Their sum is 1004+996=2000, and their difference is 1004-996=8. Therefore, \\(1004^2-996^2=2000\\times8=16000\\). Option A, 8000, is too small and can result from using only part of the calculation. Option B is correct because it uses both the sum and the difference. Option C, 2000, is only the sum, not the required product. Option D, 4000, does not follow from the identity and may come from an incorrect operation involving the difference. This identity is useful because it avoids calculating two large squares separately. Memory cue: square difference means sum multiplied by difference.

Use the identity \((a+b)^2=a^2+2ab+b^2\) with \(205=200+5\). Thus, \((205)^2=200^2+2\times200\times5+5^2=40000+2000+25=42025\). In 41025, the middle term \(2ab\) is not calculated correctly. Exam tip: while squaring a binomial, always include the \(2ab\) term.

Let \(A=6x\) and rewrite as \((A+1)^2-(A-1)^2\). Using \( (u^2-v^2)=(u-v)(u+v) \): \((A+1)^2-(A-1)^2=[(A+1)-(A-1)]\,[(A+1)+(A-1)]\). First bracket: \((A+1)-(A-1)=2\). Second bracket: \((A+1)+(A-1)=2A\). So the result is \(2\cdot2A=4A=4\cdot6x=24x\).

The identity is \(a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\), so A is correct. In B, the signs of the mixed terms are wrong. Exam tip: first check the factor \(a+b+c\).

\(x^2-49=x^2-7^2\) is a difference of two squares. Using \(a^2-b^2=(a-b)(a+b)\), it becomes \((x-7)(x+7)\). \(x^2+49\) is a sum of squares, not this identity. Exam tip: check for two perfect squares separated by a minus sign.

Expand using \((x-y)^3=x^3-3x^2y+3xy^2-y^3\). Let \(x=3a\) and \(y=b\). Then the \(a^2b\) term comes from \(-3x^2y=-3(3a)^2(b)=-3\cdot 9a^2b=-27a^2b\). So the coefficient of \(a^2b\) is \(-27\).

Use the cube identity \\(a+b)^3=a^3+3a^2b+3ab^2+b^3\\) and \\(a-b)^3=a^3-3a^2b+3ab^2-b^3\\). Put \\(a=x\\) and \\(b=4\\). Their difference is \\(6a^2b+2b^3\\), so it becomes \\(6x^2(4)+2(4^3)=24x^2+128\\). The \\(x^3\\) terms cancel, and the terms containing only \\(x\\) also cancel.

Thus option A is correct. Direct expansion gives \\(x^3+12x^2+48x+64\\) minus \\(x^3-12x^2+48x-64\\), which leaves \\(24x^2+128\\). Option B has only half of the quadratic contribution, option C incorrectly leaves an \\(x^3\\) term, and option D includes a linear term that should cancel.

In \(x^2+10x+25\), the first and last terms are \(x^2\) and \(5^2\). The middle term is \(2\times x\times5=10x\), so it equals \((x+5)^2\). Exam tip: check whether the middle term is \(2ab\).

Use the identity \(a^3-b^3=(a-b)(a^2+ab+b^2)\). Here \(64x^3=(4x)^3\) and \(27y^3=(3y)^3\). So \(64x^3-27y^3=(4x-3y)\big((4x)^2+(4x)(3y)+(3y)^2\big)\). Compute: \((4x)^2=16x^2\), \((4x)(3y)=12xy\), \((3y)^2=9y^2\). Hence \(64x^3-27y^3=(4x-3y)(16x^2+12xy+9y^2)\).

In \(m^2-14m+49\), the first and last terms are \(m^2\) and \(7^2\), while the middle term is \(-2\times m\times7=-14m\). Hence it is \((m-7)^2\). Exam tip: check both the middle-term sign and twice-product condition.

Using the identity \((x+y)^2=x^2+y^2+2xy\), we get \(x^2+y^2=(x+y)^2-2xy\). Therefore, \(x^2+y^2=13^2-2(40)=169-80=89\). Option 169 is only \((x+y)^2\); subtracting \(2xy\) is necessary. Exam tip: when the sum and product are given, start with \((x+y)^2-2xy\).