Algebraic Identities Questions in English

(A) To evaluate $103 \times 97$,we can use the algebraic identity $(a + b)(a - b) = a^2 - b^2$.
We can write $103$ as $(100 + 3)$ and $97$ as $(100 - 3)$.
Substituting these values into the identity:
$103 \times 97 = (100 + 3)(100 - 3)$
$= 100^2 - 3^2$
$= 10000 - 9$
$= 9991$.

(B) To evaluate $205 \times 195$,we can use the algebraic identity $(a + b)(a - b) = a^2 - b^2$.
We can write $205$ as $(200 + 5)$ and $195$ as $(200 - 5)$.
So,$205 \times 195 = (200 + 5)(200 - 5)$.
Applying the identity,we get $200^2 - 5^2$.
$200^2 = 40000$ and $5^2 = 25$.
Therefore,$40000 - 25 = 39975$.

(C) To evaluate $77 \times 83$,we can use the algebraic identity $(a - b)(a + b) = a^2 - b^2$.
We can write $77$ as $(80 - 3)$ and $83$ as $(80 + 3)$.
Substituting these into the identity:
$77 \times 83 = (80 - 3)(80 + 3)$
$= 80^2 - 3^2$
$= 6400 - 9$
$= 6391$.

(D) To evaluate $153 \times 147$,we can use the algebraic identity $(a + b)(a - b) = a^2 - b^2$.
We can express the numbers as:
$153 = 150 + 3$
$147 = 150 - 3$
Now,substitute these into the identity:
$(150 + 3)(150 - 3) = 150^2 - 3^2$
Calculate the squares:
$150^2 = 22500$
$3^2 = 9$
Subtract the values:
$22500 - 9 = 22491$
Therefore,$153 \times 147 = 22491$.

(N/A) To expand $(2x + 3y + 5)^2$,we use the algebraic identity: $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$.
Here,$a = 2x$,$b = 3y$,and $c = 5$.
Substituting these values into the identity:
$(2x + 3y + 5)^2 = (2x)^2 + (3y)^2 + (5)^2 + 2(2x)(3y) + 2(3y)(5) + 2(5)(2x)$.
Calculating each term:
$(2x)^2 = 4x^2$
$(3y)^2 = 9y^2$
$(5)^2 = 25$
$2(2x)(3y) = 12xy$
$2(3y)(5) = 30y$
$2(5)(2x) = 20x$
Combining these results,we get: $4x^2 + 9y^2 + 25 + 12xy + 30y + 20x$.

To expand the expression $\left(\frac{x}{2}+\frac{2 y}{3}-\frac{3 z}{4}\right)^{2}$,we use the algebraic identity $(a+b+c)^{2} = a^{2}+b^{2}+c^{2}+2ab+2bc+2ca$.
Here,$a = \frac{x}{2}$,$b = \frac{2y}{3}$,and $c = -\frac{3z}{4}$.
Substituting these values into the identity:
$= \left(\frac{x}{2}\right)^{2} + \left(\frac{2y}{3}\right)^{2} + \left(-\frac{3z}{4}\right)^{2} + 2\left(\frac{x}{2}\right)\left(\frac{2y}{3}\right) + 2\left(\frac{2y}{3}\right)\left(-\frac{3z}{4}\right) + 2\left(-\frac{3z}{4}\right)\left(\frac{x}{2}\right)$
$= \frac{x^{2}}{4} + \frac{4y^{2}}{9} + \frac{9z^{2}}{16} + \frac{2xy}{3} - yz - \frac{3zx}{4}$

(A) To expand $(5x - 7y - z)^2$,we use the algebraic identity: $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$.
Here,$a = 5x$,$b = -7y$,and $c = -z$.
Substituting these values into the identity:
$(5x - 7y - z)^2 = (5x)^2 + (-7y)^2 + (-z)^2 + 2(5x)(-7y) + 2(-7y)(-z) + 2(-z)(5x)$.
Calculating each term:
$(5x)^2 = 25x^2$
$(-7y)^2 = 49y^2$
$(-z)^2 = z^2$
$2(5x)(-7y) = -70xy$
$2(-7y)(-z) = 14yz$
$2(-z)(5x) = -10zx$
Combining these,we get: $25x^2 + 49y^2 + z^2 - 70xy + 14yz - 10zx$.

(A) To expand $(a - 2b + 7c)^2$,we use the algebraic identity: $(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$.
Here,$x = a$,$y = -2b$,and $z = 7c$.
Substituting these values into the identity:
$(a - 2b + 7c)^2 = (a)^2 + (-2b)^2 + (7c)^2 + 2(a)(-2b) + 2(-2b)(7c) + 2(7c)(a)$.
Calculating each term:
$= a^2 + 4b^2 + 49c^2 - 4ab - 28bc + 14ca$.

(N/A) The given expression is $x^{2}+4 y^{2}+25 z^{2}+4 x y-20 y z-10 z x$.
We can write this in the form of the algebraic identity $(a+b+c)^{2} = a^{2}+b^{2}+c^{2}+2ab+2bc+2ca$.
Here,$a = x$,$b = 2y$,and $c = -5z$.
Substituting these values into the identity:
$x^{2} + (2y)^{2} + (-5z)^{2} + 2(x)(2y) + 2(2y)(-5z) + 2(-5z)(x)$
$= x^{2} + 4y^{2} + 25z^{2} + 4xy - 20yz - 10zx$.
Thus,the expression is equivalent to $(x+2y-5z)^{2}$.
Therefore,the factors are $(x+2y-5z)(x+2y-5z)$.

(A) We use the algebraic identity: $(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$.
Given expression: $4x^2 + 9y^2 + 49z^2 - 12xy + 42yz - 28zx$.
This can be written as: $(2x)^2 + (-3y)^2 + (-7z)^2 + 2(2x)(-3y) + 2(-3y)(-7z) + 2(-7z)(2x)$.
Here,$a = 2x$,$b = -3y$,and $c = -7z$.
Therefore,the expression is equal to $(2x - 3y - 7z)^2$,which is $(2x - 3y - 7z)(2x - 3y - 7z)$.

(N/A) The given expression is $25 x^{2}+9 y^{2}+64+30 x y-48 y-80 x$.
This can be rewritten as $(5 x)^{2}+(3 y)^{2}+(-8)^{2}+2(5 x)(3 y)+2(3 y)(-8)+2(5 x)(-8)$.
This expression is in the form of the algebraic identity $a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a = (a+b+c)^{2}$.
Here,$a = 5 x$,$b = 3 y$,and $c = -8$.
Therefore,the expression becomes $(5 x+3 y-8)^{2}$,which is $(5 x+3 y-8)(5 x+3 y-8)$.

(N/A) The given expression is of the form $a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca = (a + b + c)^{2}$.
Comparing the given expression $x^{2} + (\frac{y}{2})^{2} + (\frac{z}{4})^{2} + 2(x)(\frac{y}{2}) + 2(\frac{y}{2})(\frac{z}{4}) + 2(\frac{z}{4})(x)$ with the identity:
Here,$a = x$,$b = \frac{y}{2}$,and $c = \frac{z}{4}$.
Thus,the expression can be written as $(x + \frac{y}{2} + \frac{z}{4})^{2}$.
Therefore,the factors are $(x + \frac{y}{2} + \frac{z}{4})(x + \frac{y}{2} + \frac{z}{4})$.

(A) To evaluate $(153)^{2}$,we can use the algebraic identity $(a + b)^{2} = a^{2} + 2ab + b^{2}$.
We can write $153$ as $(150 + 3)$.
So,$(153)^{2} = (150 + 3)^{2}$.
Using the identity,we get:
$(150)^{2} + 2(150)(3) + (3)^{2}$
$= 22500 + 900 + 9$
$= 23409$.
Therefore,the correct value is $23409$.

(B) To evaluate $(215)^{2}$,we can use the identity $(a + b)^{2} = a^{2} + 2ab + b^{2}$.
We can write $215$ as $(200 + 15)$.
$(200 + 15)^{2} = (200)^{2} + 2(200)(15) + (15)^{2}$.
$= 40000 + 6000 + 225$.
$= 46225$.
Thus,the correct option is $B$.

(C) To evaluate $(421)^{2}$,we can use the algebraic identity $(a + b)^{2} = a^{2} + 2ab + b^{2}$.
We can write $421$ as $(400 + 21)$.
Applying the identity: $(400 + 21)^{2} = (400)^{2} + 2(400)(21) + (21)^{2}$.
$(400)^{2} = 160000$.
$2 \times 400 \times 21 = 800 \times 21 = 16800$.
$(21)^{2} = 441$.
Adding these values together: $160000 + 16800 + 441 = 177241$.
Therefore,$(421)^{2} = 177241$.

(A) To evaluate $(555)^{2}$,we can use the algebraic identity $(a + b)^{2} = a^{2} + 2ab + b^{2}$.
We can write $555$ as $(500 + 55)$.
$(555)^{2} = (500 + 55)^{2}$
$= (500)^{2} + 2(500)(55) + (55)^{2}$
$= 250000 + 55000 + 3025$
$= 305000 + 3025$
$= 308025$.

(N/A) To expand $(3a + 5b)^3$,we use the algebraic identity $(x + y)^3 = x^3 + y^3 + 3xy(x + y)$.
Here,$x = 3a$ and $y = 5b$.
Substituting these values into the identity:
$(3a + 5b)^3 = (3a)^3 + (5b)^3 + 3(3a)(5b)(3a + 5b)$
$= 27a^3 + 125b^3 + 45ab(3a + 5b)$
$= 27a^3 + 125b^3 + 135a^2b + 225ab^2$.

To expand the expression $(2x + 7)^3$,we use the algebraic identity: $(a + b)^3 = a^3 + b^3 + 3a^2b + 3ab^2$.
Here,$a = 2x$ and $b = 7$.
Substituting these values into the identity:
$(2x + 7)^3 = (2x)^3 + (7)^3 + 3(2x)^2(7) + 3(2x)(7)^2$
$= 8x^3 + 343 + 3(4x^2)(7) + 3(2x)(49)$
$= 8x^3 + 343 + 84x^2 + 294x$
Rearranging the terms in descending order of powers:
$= 8x^3 + 84x^2 + 294x + 343$.

(N/A) To expand $(4x - 3y)^{3}$,we use the algebraic identity:
$(a - b)^{3} = a^{3} - b^{3} - 3ab(a - b) = a^{3} - 3a^{2}b + 3ab^{2} - b^{3}$
Here,$a = 4x$ and $b = 3y$.
Substituting these values into the identity:
$(4x - 3y)^{3} = (4x)^{3} - 3(4x)^{2}(3y) + 3(4x)(3y)^{2} - (3y)^{3}$
$= 64x^{3} - 3(16x^{2})(3y) + 3(4x)(9y^{2}) - 27y^{3}$
$= 64x^{3} - 144x^{2}y + 108xy^{2} - 27y^{3}$

(N/A) To expand $(7x - 4y)^{3}$,we use the algebraic identity: $(a - b)^{3} = a^{3} - b^{3} - 3a^{2}b + 3ab^{2}$.
Here,$a = 7x$ and $b = 4y$.
Substituting these values into the identity:
$(7x - 4y)^{3} = (7x)^{3} - (4y)^{3} - 3(7x)^{2}(4y) + 3(7x)(4y)^{2}$
$= 343x^{3} - 64y^{3} - 3(49x^{2})(4y) + 3(7x)(16y^{2})$
$= 343x^{3} - 64y^{3} - 588x^{2}y + 336xy^{2}$.

(N/A) The given expression is $8x^3 + 343y^3 + 84x^2y + 294xy^2$.
We can rewrite this as $(2x)^3 + (7y)^3 + 3(2x)^2(7y) + 3(2x)(7y)^2$.
This expression is in the form of the algebraic identity $a^3 + b^3 + 3a^2b + 3ab^2 = (a + b)^3$,where $a = 2x$ and $b = 7y$.
Substituting these values,we get $(2x + 7y)^3$.
Thus,the factorised form is $(2x + 7y)(2x + 7y)(2x + 7y)$.

(A) The given expression is $x^{3} - 125y^{3} - 15x^{2}y + 75xy^{2}$.
We can rewrite this as $x^{3} - (5y)^{3} - 3(x^{2})(5y) + 3(x)(5y)^{2}$.
This expression is in the form of the algebraic identity $(a - b)^{3} = a^{3} - b^{3} - 3a^{2}b + 3ab^{2}$,where $a = x$ and $b = 5y$.
Therefore,the expression simplifies to $(x - 5y)^{3}$,which can be written as $(x - 5y)(x - 5y)(x - 5y)$.

(A) The given expression is $64 x^{3} + 125 y^{3} + 240 x^{2} y + 300 x y^{2}$.
We can rewrite this expression as $(4x)^{3} + (5y)^{3} + 3(4x)^{2}(5y) + 3(4x)(5y)^{2}$.
This is in the form of the algebraic identity $a^{3} + b^{3} + 3a^{2}b + 3ab^{2} = (a + b)^{3}$,where $a = 4x$ and $b = 5y$.
Substituting these values,we get $(4x + 5y)^{3}$.
Therefore,the factorised form is $(4x + 5y)(4x + 5y)(4x + 5y)$.

(A) The given expression is $27 x^{3}-8 y^{3}-54 x^{2} y+36 x y^{2}$.
This can be rewritten as $(3x)^3 - (2y)^3 - 3(3x)^2(2y) + 3(3x)(2y)^2$.
Recall the algebraic identity: $(a - b)^3 = a^3 - b^3 - 3a^2b + 3ab^2$.
Here,$a = 3x$ and $b = 2y$.
Substituting these values into the identity,we get $(3x - 2y)^3$.
Thus,the factorised form is $(3x - 2y)(3x - 2y)(3x - 2y)$.

(A) To evaluate $(995)^{3}$,we can express $995$ as $(1000 - 5)$.
Using the algebraic identity $(a - b)^{3} = a^{3} - b^{3} - 3ab(a - b)$:
Here,$a = 1000$ and $b = 5$.
$(1000 - 5)^{3} = (1000)^{3} - (5)^{3} - 3(1000)(5)(1000 - 5)$
$= 1,000,000,000 - 125 - 15000(995)$
$= 1,000,000,000 - 125 - 14,925,000$
$= 1,000,000,000 - 14,925,125$
$= 985,074,875$.

(B) To evaluate $(105)^{3}$,we can express $105$ as $(100 + 5)$.
Using the algebraic identity $(a + b)^{3} = a^{3} + b^{3} + 3ab(a + b)$:
Here,$a = 100$ and $b = 5$.
$(100 + 5)^{3} = (100)^{3} + (5)^{3} + 3(100)(5)(100 + 5)$
$= 1000000 + 125 + 1500(105)$
$= 1000000 + 125 + 157500$
$= 1157625$

(N/A) The given expression is of the form $a^{3}+b^{3}+c^{3}-3 a b c$,where $a = 2x$,$b = 3y$,and $c = 5z$.
Using the algebraic identity $a^{3}+b^{3}+c^{3}-3 a b c = (a+b+c)(a^{2}+b^{2}+c^{2}-a b-b c-c a)$:
Substitute $a = 2x$,$b = 3y$,and $c = 5z$ into the identity:
$8 x^{3}+27 y^{3}+125 z^{3}-3(2x)(3y)(5z) = (2x+3y+5z)((2x)^{2}+(3y)^{2}+(5z)^{2}-(2x)(3y)-(3y)(5z)-(5z)(2x))$
$= (2x+3y+5z)(4x^{2}+9y^{2}+25z^{2}-6xy-15yz-10zx)$.

(A) We use the algebraic identity: $a^{3}+b^{3}+c^{3}-3 a b c = (a+b+c)(a^{2}+b^{2}+c^{2}-a b-b c-c a)$.
Given expression: $27 x^{3}-y^{3}+64 z^{3}+36 x y z = (3 x)^{3}+(-y)^{3}+(4 z)^{3}-3(3 x)(-y)(4 z)$.
Here,$a = 3 x$,$b = -y$,and $c = 4 z$.
Substituting these values into the identity:
$= (3 x-y+4 z)((3 x)^{2}+(-y)^{2}+(4 z)^{2}-(3 x)(-y)-(-y)(4 z)-(4 z)(3 x))$.
$= (3 x-y+4 z)(9 x^{2}+y^{2}+16 z^{2}+3 x y+4 y z-12 z x)$.

(B) The given expression is $x^{3}-8 y^{3}-27-18 x y$.
This can be rewritten as $x^{3} + (-2y)^{3} + (-3)^{3} - 3(x)(-2y)(-3)$.
We use the algebraic identity $a^{3} + b^{3} + c^{3} - 3abc = (a+b+c)(a^{2} + b^{2} + c^{2} - ab - bc - ca)$.
Here,$a = x$,$b = -2y$,and $c = -3$.
Substituting these values into the identity:
$= (x - 2y - 3)(x^{2} + (-2y)^{2} + (-3)^{2} - (x)(-2y) - (-2y)(-3) - (-3)(x))$
$= (x - 2y - 3)(x^{2} + 4y^{2} + 9 + 2xy - 6y + 3x)$.

(A) The given expression is of the form $a^{3} + b^{3} + c^{3} - 3abc$,where $a = 2x$,$b = 5y$,and $c = 7$.
We know the identity: $a^{3} + b^{3} + c^{3} - 3abc = (a + b + c)(a^{2} + b^{2} + c^{2} - ab - bc - ca)$.
Here,$3abc = 3(2x)(5y)(7) = 210xy$.
Substituting the values into the identity:
$8x^{3} + 125y^{3} + 343 - 210xy = (2x + 5y + 7)((2x)^{2} + (5y)^{2} + (7)^{2} - (2x)(5y) - (5y)(7) - (7)(2x))$.
Simplifying the terms:
$= (2x + 5y + 7)(4x^{2} + 25y^{2} + 49 - 10xy - 35y - 14x)$.

(A) We use the algebraic identity: If $a+b+c=0$,then $a^3+b^3+c^3=3abc$.
Let $a = (x-2y)$,$b = (2y-3z)$,and $c = (3z-x)$.
Now,calculate the sum $a+b+c$:
$a+b+c = (x-2y) + (2y-3z) + (3z-x)$
$a+b+c = x - x - 2y + 2y - 3z + 3z = 0$.
Since the sum is $0$,the expression $(x-2y)^3 + (2y-3z)^3 + (3z-x)^3$ simplifies to $3abc$.
Therefore,the factorised form is $3(x-2y)(2y-3z)(3z-x)$.

(D) We use the algebraic identity: If $x + y + z = 0$,then $x^3 + y^3 + z^3 = 3xyz$.
Given expression: $(31)^3 + (-16)^3 + (-15)^3$.
Let $x = 31$,$y = -16$,and $z = -15$.
Check the sum: $x + y + z = 31 + (-16) + (-15) = 31 - 31 = 0$.
Since the sum is $0$,the value of the expression is $3xyz$.
Value $= 3 \times (31) \times (-16) \times (-15)$.
Value $= 3 \times 31 \times 240$.
Value $= 93 \times 240 = 22320$.

(A) We know the algebraic identity: If $x + y + z = 0$,then $x^{3} + y^{3} + z^{3} = 3xyz$.
Given expression: $(14)^{3} + (27)^{3} - (41)^{3}$.
This can be rewritten as: $(14)^{3} + (27)^{3} + (-41)^{3}$.
Let $x = 14$,$y = 27$,and $z = -41$.
Check the sum: $x + y + z = 14 + 27 + (-41) = 41 - 41 = 0$.
Since the sum is $0$,the expression equals $3xyz$.
Value $= 3 \times (14) \times (27) \times (-41)$.
Value $= 3 \times 378 \times (-41) = 1134 \times (-41) = -46494$.

(B) Let $x = \frac{1}{2}$,$y = \frac{1}{3}$,and $z = -\frac{5}{6}$.
First,check the sum of $x, y,$ and $z$:
$x + y + z = \frac{1}{2} + \frac{1}{3} - \frac{5}{6} = \frac{3+2-5}{6} = \frac{0}{6} = 0$.
We know the algebraic identity: If $x + y + z = 0$,then $x^3 + y^3 + z^3 = 3xyz$.
Substituting the values:
$x^3 + y^3 + z^3 = 3 \times \left(\frac{1}{2}\right) \times \left(\frac{1}{3}\right) \times \left(-\frac{5}{6}\right)$.
$= 3 \times \left(\frac{1}{6}\right) \times \left(-\frac{5}{6}\right) = 1 \times \left(-\frac{5}{6}\right) = -\frac{5}{6}$.
Wait,re-calculating: $3 \times \frac{1}{2} \times \frac{1}{3} \times (-\frac{5}{6}) = 1 \times (-\frac{5}{6}) = -\frac{5}{6}$.
Correction: The expression is $3 \times \frac{1}{2} \times \frac{1}{3} \times (-\frac{5}{6}) = \frac{3}{6} \times (-\frac{5}{6}) = \frac{1}{2} \times (-\frac{5}{6}) = -\frac{5}{12}$.

(C) Let $a = 0.2$,$b = -0.3$,and $c = 0.1$.
We observe that $a + b + c = 0.2 + (-0.3) + 0.1 = 0.3 - 0.3 = 0$.
We know the algebraic identity: If $a + b + c = 0$,then $a^{3} + b^{3} + c^{3} = 3abc$.
Substituting the values:
$(0.2)^{3} + (-0.3)^{3} + (0.1)^{3} = 3(0.2)(-0.3)(0.1)$.
Calculating the product: $3 \times 0.2 = 0.6$; $0.6 \times (-0.3) = -0.18$; $-0.18 \times 0.1 = -0.018$.
Thus,$(0.2)^{3} - (0.3)^{3} + (0.1)^{3} = -0.018$.

(A) The area of a square is given by the formula: $\text{Area} = (\text{side})^2$.
Given, $\text{Area} = 9x^2 + 30x + 25$.
We can factorize this quadratic expression:
$9x^2 + 30x + 25 = (3x)^2 + 2(3x)(5) + (5)^2$.
Using the algebraic identity $(a + b)^2 = a^2 + 2ab + b^2$, where $a = 3x$ and $b = 5$, we get:
$9x^2 + 30x + 25 = (3x + 5)^2$.
Since $\text{Area} = (\text{side})^2$, the length of the side is $(3x + 5)$ units.

(A) Given the equation $x+y = -4$.
Cubing both sides,we get $(x+y)^3 = (-4)^3$.
Using the identity $(x+y)^3 = x^3 + y^3 + 3xy(x+y)$,we have $x^3 + y^3 + 3xy(x+y) = -64$.
Substitute the value of $(x+y) = -4$ into the equation:
$x^3 + y^3 + 3xy(-4) = -64$.
$x^3 + y^3 - 12xy = -64$.
Adding $64$ to both sides,we get $x^3 + y^3 - 12xy + 64 = -64 + 64$.
Therefore,$x^3 + y^3 - 12xy + 64 = 0$.

(B) Given the equation $x = 2y + 6$,we can rewrite it as $x - 2y = 6$.
To find the value of $x^3 - 8y^3 - 36xy - 216$,we can express it as $x^3 + (-2y)^3 + (-6)^3 - 3(x)(-2y)(-6)$.
Recall the algebraic identity: $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$.
Alternatively,consider the identity $(a + b + c)^3 = a^3 + b^3 + c^3 + 3(a + b)(b + c)(c + a)$.
Let $a = x$,$b = -2y$,and $c = -6$.
Then $a + b + c = x - 2y - 6$. Since $x - 2y = 6$,we have $a + b + c = 6 - 6 = 0$.
If $a + b + c = 0$,then $a^3 + b^3 + c^3 = 3abc$.
Substituting the values: $x^3 + (-2y)^3 + (-6)^3 = 3(x)(-2y)(-6)$.
$x^3 - 8y^3 - 216 = 36xy$.
Rearranging the terms: $x^3 - 8y^3 - 36xy - 216 = 0$.

(B) To calculate $85 \times 75$,we can use the algebraic identity $(a - b)(a + b) = a^2 - b^2$.
We can write $85$ as $(80 + 5)$ and $75$ as $(80 - 5)$.
Thus,$85 \times 75 = (80 + 5)(80 - 5)$.
Using the identity $a^2 - b^2$,where $a = 80$ and $b = 5$:
$= 80^2 - 5^2$
$= 6400 - 25$
$= 6375$.

(C) To factorize the quadratic expression $4x^{2} - 20x + 25$,we observe that it follows the algebraic identity $(a - b)^{2} = a^{2} - 2ab + b^{2}$.
Here,$a^{2} = 4x^{2} \implies a = 2x$.
Also,$b^{2} = 25 \implies b = 5$.
Checking the middle term: $-2ab = -2(2x)(5) = -20x$,which matches the given expression.
Therefore,$4x^{2} - 20x + 25 = (2x - 5)^{2}$.

(B) The given expression is in the form of the algebraic identity $(a+b)(a-b) = a^{2}-b^{2}$.
Here,$a = 5x$ and $b = 3$.
Applying the identity:
$(5x+3)(5x-3) = (5x)^{2} - (3)^{2}$
$= 25x^{2} - 9$.

(B) The given expression is $49x^2 - 121$.
This can be written in the form of $a^2 - b^2$ as follows:
$49x^2 - 121 = (7x)^2 - (11)^2$.
Using the algebraic identity $a^2 - b^2 = (a + b)(a - b)$,where $a = 7x$ and $b = 11$,we get:
$(7x)^2 - (11)^2 = (7x + 11)(7x - 11)$.
Therefore,the factors are $(7x + 11)(7x - 11)$.

(D) Given the equation: $(5x - 3)^2 = 25x^2 + kx + 9$.
We use the algebraic identity $(a - b)^2 = a^2 - 2ab + b^2$.
Here,$a = 5x$ and $b = 3$.
Expanding the left side: $(5x - 3)^2 = (5x)^2 - 2(5x)(3) + (3)^2$.
$(5x - 3)^2 = 25x^2 - 30x + 9$.
Comparing this with the given expression $25x^2 + kx + 9$,we get:
$25x^2 - 30x + 9 = 25x^2 + kx + 9$.
By comparing the coefficients of $x$,we find $k = -30$.

(A) To find the value of $k$,we expand the left side of the equation:
$(2x + 3)(3x - 1) = 2x(3x) + 2x(-1) + 3(3x) + 3(-1)$
$= 6x^2 - 2x + 9x - 3$
$= 6x^2 + 7x - 3$
Comparing this with the given equation $6x^2 + kx - 3$,we can see that the coefficient of $x$ is $k$ on the right side and $7$ on the expanded left side.
Therefore,$k = 7$.