Mix Examples - Polynomials Questions in English

(A) To factorise the quadratic expression $9 x^{2}-21 x y+10 y^{2}$,we use the splitting the middle term method.
We need to find two numbers whose product is $9 \times 10 = 90$ and whose sum is $-21$.
The two numbers are $-15$ and $-6$,since $(-15) \times (-6) = 90$ and $(-15) + (-6) = -21$.
Now,rewrite the middle term $-21 x y$ as $-15 x y - 6 x y$:
$9 x^{2} - 15 x y - 6 x y + 10 y^{2}$
Group the terms:
$(9 x^{2} - 15 x y) - (6 x y - 10 y^{2})$
Factor out the common terms from each group:
$3 x(3 x - 5 y) - 2 y(3 x - 5 y)$
Finally,factor out the common binomial $(3 x - 5 y)$:
$(3 x - 5 y)(3 x - 2 y)$

(A) To factorise the quadratic expression $49 x^{2}-35 x+6$,we use the splitting the middle term method.
We need to find two numbers such that their product is $49 \times 6 = 294$ and their sum is $-35$.
The two numbers are $-21$ and $-14$,since $(-21) \times (-14) = 294$ and $(-21) + (-14) = -35$.
Now,rewrite the middle term:
$49 x^{2}-21 x-14 x+6$
Group the terms:
$(49 x^{2}-21 x) - (14 x-6)$
Factor out the common terms from each group:
$7 x(7 x-3) - 2(7 x-3)$
Finally,factor out the common binomial $(7 x-3)$:
$(7 x-3)(7 x-2)$

(C) To evaluate $103 \times 105$ without direct multiplication,we can use the algebraic identity: $(x + a)(x + b) = x^2 + (a + b)x + ab$.
Here,we can write $103 = (100 + 3)$ and $105 = (100 + 5)$.
Comparing this with the identity,we have $x = 100$,$a = 3$,and $b = 5$.
Substituting these values into the identity:
$(100 + 3)(100 + 5) = (100)^2 + (3 + 5)(100) + (3 \times 5)$
$= 10000 + (8 \times 100) + 15$
$= 10000 + 800 + 15$
$= 10815$.

(D) To evaluate $84 \times 79$ without direct multiplication,we can express the numbers as binomials.
We can write $84$ as $(80 + 4)$ and $79$ as $(80 - 1)$.
Using the algebraic identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,where $x = 80$,$a = 4$,and $b = -1$:
$(80 + 4)(80 - 1) = 80^2 + (4 - 1)80 + (4 \times -1)$
$= 6400 + (3 \times 80) - 4$
$= 6400 + 240 - 4$
$= 6640 - 4$
$= 6636$.

(A) To evaluate $76 \times 82$ without direct multiplication,we can express the numbers as $(79 - 3)$ and $(79 + 3)$.
Using the algebraic identity $(a - b)(a + b) = a^2 - b^2$,where $a = 79$ and $b = 3$:
$76 \times 82 = (79 - 3)(79 + 3)$
$= 79^2 - 3^2$
$= 6241 - 9$
$= 6232$.

(B) To evaluate $88 \times 86$ without direct multiplication,we can express the numbers as $(90 - 2) \times (90 - 4)$.
Using the algebraic identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,where $x = 90$,$a = -2$,and $b = -4$:
$(90 - 2)(90 - 4) = 90^2 + (-2 - 4) \times 90 + (-2 \times -4)$
$= 8100 + (-6 \times 90) + 8$
$= 8100 - 540 + 8$
$= 7560 + 8$
$= 7568$.

(A) To expand the expression $(3x + 5)^2$,we use the algebraic identity $(a + b)^2 = a^2 + 2ab + b^2$.
Here,$a = 3x$ and $b = 5$.
Substituting these values into the identity:
$(3x + 5)^2 = (3x)^2 + 2(3x)(5) + (5)^2$
$= 9x^2 + 30x + 25$.

(B) To expand the expression $(6x - 7)^2$,we use the algebraic identity $(a - b)^2 = a^2 - 2ab + b^2$.
Here,$a = 6x$ and $b = 7$.
Substituting these values into the identity:
$(6x - 7)^2 = (6x)^2 - 2(6x)(7) + (7)^2$
$= 36x^2 - 84x + 49$.
Thus,the expanded form is $36x^2 - 84x + 49$.

(B) To expand $(2a + 3b)^2$,we use the algebraic identity $(x + y)^2 = x^2 + 2xy + y^2$.
Here,$x = 2a$ and $y = 3b$.
Substituting these values into the identity:
$(2a + 3b)^2 = (2a)^2 + 2(2a)(3b) + (3b)^2$
$= 4a^2 + 12ab + 9b^2$.

(A) To expand the expression $\left(\frac{x}{2}-\frac{2}{5}\right)^{2}$,we use the algebraic identity $(a-b)^{2} = a^{2} - 2ab + b^{2}$.
Here,$a = \frac{x}{2}$ and $b = \frac{2}{5}$.
Substituting these values into the identity:
$\left(\frac{x}{2}-\frac{2}{5}\right)^{2} = \left(\frac{x}{2}\right)^{2} - 2\left(\frac{x}{2}\right)\left(\frac{2}{5}\right) + \left(\frac{2}{5}\right)^{2}$
$= \frac{x^{2}}{4} - \frac{4x}{10} + \frac{4}{25}$
$= \frac{x^{2}}{4} - \frac{2x}{5} + \frac{4}{25}$.

(A) To expand the expression $\left(\frac{2x}{3} + \frac{3y}{4}\right)^{2}$,we use the algebraic identity $(a + b)^{2} = a^{2} + 2ab + b^{2}$.
Here,$a = \frac{2x}{3}$ and $b = \frac{3y}{4}$.
Substituting these values into the identity:
$\left(\frac{2x}{3} + \frac{3y}{4}\right)^{2} = \left(\frac{2x}{3}\right)^{2} + 2\left(\frac{2x}{3}\right)\left(\frac{3y}{4}\right) + \left(\frac{3y}{4}\right)^{2}$
$= \frac{4x^{2}}{9} + 2\left(\frac{6xy}{12}\right) + \frac{9y^{2}}{16}$
$= \frac{4}{9}x^{2} + 2\left(\frac{xy}{2}\right) + \frac{9}{16}y^{2}$
$= \frac{4}{9}x^{2} + xy + \frac{9}{16}y^{2}$.

(A) To expand the expression $(x - \frac{1}{2})^{2}$,we use the algebraic identity $(a - b)^{2} = a^{2} - 2ab + b^{2}$.
Here,$a = x$ and $b = \frac{1}{2}$.
Substituting these values into the identity:
$(x - \frac{1}{2})^{2} = (x)^{2} - 2(x)(\frac{1}{2}) + (\frac{1}{2})^{2}$.
Simplifying the terms:
$= x^{2} - x + \frac{1}{4}$.

(B) The given expression is $16 x^{2}-40 x y+25 y^{2}$.
We can rewrite this expression as $(4 x)^{2}-2(4 x)(5 y)+(5 y)^{2}$.
This is in the form of the algebraic identity $a^{2}-2 a b+b^{2}=(a-b)^{2}$,where $a=4 x$ and $b=5 y$.
Substituting these values,we get $(4 x-5 y)^{2}$.

(A) To factorise the expression $9x^{2} + 42x + 49$,we observe that it follows the algebraic identity $(a + b)^{2} = a^{2} + 2ab + b^{2}$.
We can rewrite the expression as:
$9x^{2} + 42x + 49 = (3x)^{2} + 2(3x)(7) + (7)^{2}$.
Here,$a = 3x$ and $b = 7$.
Therefore,the expression simplifies to $(3x + 7)^{2}$.

(A) The given expression is $\frac{4x^2}{9} - \frac{x}{3} + \frac{1}{16}$.
This can be rewritten in the form $a^2 - 2ab + b^2 = (a - b)^2$.
Here,$a^2 = \frac{4x^2}{9} \implies a = \frac{2x}{3}$.
And $b^2 = \frac{1}{16} \implies b = \frac{1}{4}$.
Checking the middle term: $-2ab = -2 \times \left(\frac{2x}{3}\right) \times \left(\frac{1}{4}\right) = -\frac{4x}{12} = -\frac{x}{3}$.
Since the middle term matches,the expression is $\left(\frac{2x}{3} - \frac{1}{4}\right)^2$.

(A) The given expression is $\frac{x^{2}}{4}+\frac{3 x y}{5}+\frac{9 y^{2}}{25}$.
This can be written in the form $a^{2} + 2ab + b^{2} = (a + b)^{2}$.
Here,$a^{2} = \frac{x^{2}}{4} \implies a = \frac{x}{2}$.
And $b^{2} = \frac{9 y^{2}}{25} \implies b = \frac{3 y}{5}$.
Now,check the middle term $2ab = 2 \times \left(\frac{x}{2}\right) \times \left(\frac{3 y}{5}\right) = \frac{3 x y}{5}$.
Since the middle term matches,the expression is $\left(\frac{x}{2}+\frac{3 y}{5}\right)^{2}$.

(A) To evaluate $(65)^{2}$,we can use the algebraic identity $(a + b)^{2} = a^{2} + 2ab + b^{2}$.
Here,we can write $65$ as $(60 + 5)$.
So,$(65)^{2} = (60 + 5)^{2}$.
Using the identity: $(60)^{2} + 2(60)(5) + (5)^{2}$.
$= 3600 + 600 + 25$.
$= 4225$.
Alternatively,for any number ending in $5$,the square is calculated as $(n5)^{2} = [n(n+1)] \times 100 + 25$.
For $65$,$n = 6$,so $[6(6+1)] \times 100 + 25 = (6 \times 7) \times 100 + 25 = 4200 + 25 = 4225$.

(B) To evaluate $(101)^{2}$,we can express $101$ as $(100 + 1)$.
Using the algebraic identity $(a + b)^{2} = a^{2} + 2ab + b^{2}$,where $a = 100$ and $b = 1$:
$(100 + 1)^{2} = (100)^{2} + 2(100)(1) + (1)^{2}$
$= 10000 + 200 + 1$
$= 10201$.

(C) To evaluate $(1002)^{2}$,we can express $1002$ as $(1000 + 2)$.
Using the algebraic identity $(a + b)^{2} = a^{2} + 2ab + b^{2}$,where $a = 1000$ and $b = 2$:
$(1000 + 2)^{2} = (1000)^{2} + 2(1000)(2) + (2)^{2}$
$= 1000000 + 4000 + 4$
$= 1004004$.

(D) To evaluate $(98)^{2}$,we can express $98$ as $(100 - 2)$.
Using the algebraic identity $(a - b)^{2} = a^{2} - 2ab + b^{2}$,where $a = 100$ and $b = 2$:
$(100 - 2)^{2} = (100)^{2} - 2(100)(2) + (2)^{2}$
$= 10000 - 400 + 4$
$= 9600 + 4$
$= 9604$.

(A) To expand the expression $(x+5y)(x-5y)$,we use the algebraic identity $(a+b)(a-b) = a^2 - b^2$.
Here,$a = x$ and $b = 5y$.
Substituting these values into the identity:
$(x+5y)(x-5y) = (x)^2 - (5y)^2$
$= x^2 - 25y^2$.

(B) The given expression is in the form of $(a + b)(a - b)$,where $a = \frac{2x}{3}$ and $b = \frac{4y}{5}$.
Using the algebraic identity $(a + b)(a - b) = a^2 - b^2$,we get:
$\left(\frac{2x}{3}\right)^2 - \left(\frac{4y}{5}\right)^2$
$= \frac{4x^2}{9} - \frac{16y^2}{25}$
$= \frac{4}{9}x^2 - \frac{16}{25}y^2$.

(N/A) To expand the expression $(3x + 7)(-3x + 7)$,we can rearrange the terms to use the algebraic identity $(a + b)(a - b) = a^2 - b^2$.
First,rewrite the expression as $(7 + 3x)(7 - 3x)$.
Here,$a = 7$ and $b = 3x$.
Applying the identity: $(7)^2 - (3x)^2 = 49 - 9x^2$.

(A) To expand the expression $(11 x+18)(11 x-18)$,we use the algebraic identity $(a+b)(a-b) = a^{2}-b^{2}$.
Here,$a = 11 x$ and $b = 18$.
Applying the identity:
$(11 x)^{2} - (18)^{2}$
$= 121 x^{2} - 324$.

(A) To factorise the expression $144 x^{2}-289 y^{2}$,we use the algebraic identity $a^{2}-b^{2}=(a-b)(a+b)$.
First,express the given terms as squares:
$144 x^{2} = (12 x)^{2}$
$289 y^{2} = (17 y)^{2}$
Now,substitute these into the identity:
$144 x^{2}-289 y^{2} = (12 x)^{2}-(17 y)^{2}$
Using the identity $a^{2}-b^{2}=(a-b)(a+b)$,where $a=12 x$ and $b=17 y$,we get:
$(12 x-17 y)(12 x+17 y)$.

(A) The given expression is $\frac{4x^2}{9} - \frac{1}{25}$.
This can be written in the form of $a^2 - b^2$,where $a^2 = \frac{4x^2}{9}$ and $b^2 = \frac{1}{25}$.
Taking the square root,we get $a = \frac{2x}{3}$ and $b = \frac{1}{5}$.
Using the algebraic identity $a^2 - b^2 = (a + b)(a - b)$,we substitute the values of $a$ and $b$.
Therefore,$\frac{4x^2}{9} - \frac{1}{25} = \left(\frac{2x}{3} + \frac{1}{5}\right)\left(\frac{2x}{3} - \frac{1}{5}\right)$.

(A) The given expression is $169 x^{2}-625$.
We can write this as $(13 x)^{2}-(25)^{2}$.
Using the algebraic identity $a^{2}-b^{2} = (a+b)(a-b)$,where $a = 13 x$ and $b = 25$,we get:
$(13 x)^{2}-(25)^{2} = (13 x+25)(13 x-25)$.

(A) To factorise $16 x^{4}-y^{4}$,we use the difference of squares identity: $a^{2}-b^{2}=(a-b)(a+b)$.
First,rewrite the expression as $(4 x^{2})^{2}-(y^{2})^{2}$.
Applying the identity,we get $(4 x^{2}-y^{2})(4 x^{2}+y^{2})$.
Now,factorise the term $(4 x^{2}-y^{2})$ further by writing it as $(2 x)^{2}-(y)^{2}$.
Using the identity again,$(4 x^{2}-y^{2}) = (2 x-y)(2 x+y)$.
Combining these,the final factorised form is $(2 x-y)(2 x+y)(4 x^{2}+y^{2})$.

(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 $(x - 5y)(x - 5y)(x - 5y)$.