Algebraic Identities Questions in English

(N/A) To verify the identity,we expand the right-hand side ($R$.$H$.$S$.).
$R.H.S. = (x+y)(x^{2}-xy+y^{2})$
$= x(x^{2}-xy+y^{2}) + y(x^{2}-xy+y^{2})$
$= (x^{3} - x^{2}y + xy^{2}) + (x^{2}y - xy^{2} + y^{3})$
By grouping the like terms,we get:
$= x^{3} + (-x^{2}y + x^{2}y) + (xy^{2} - xy^{2}) + y^{3}$
$= x^{3} + 0 + 0 + y^{3}$
$= x^{3} + y^{3} = L.H.S.$
Hence,the identity is verified.

(N/A) To verify the identity,we start with the Right Hand Side ($R$.$H$.$S$.):
$R.H.S. = (x-y)(x^{2}+xy+y^{2})$
Distribute the terms:
$= x(x^{2}+xy+y^{2}) - y(x^{2}+xy+y^{2})$
$= (x^{3} + x^{2}y + xy^{2}) - (x^{2}y + xy^{2} + y^{3})$
$= x^{3} + x^{2}y + xy^{2} - x^{2}y - xy^{2} - y^{3}$
Cancel the like terms with opposite signs ($x^{2}y - x^{2}y = 0$ and $xy^{2} - xy^{2} = 0$):
$= x^{3} - y^{3}$
$= L.H.S.$
Since $L.H.S. = R.H.S.$,the identity is verified.

(A) Using the algebraic identity $x^{3} - y^{3} = (x - y)(x^{2} + xy + y^{2})$,we can factorise the expression.
Given expression: $64m^{3} - 343n^{3}$
This can be written as: $(4m)^{3} - (7n)^{3}$
Applying the identity where $x = 4m$ and $y = 7n$:
$(4m - 7n)((4m)^{2} + (4m)(7n) + (7n)^{2})$
$= (4m - 7n)(16m^{2} + 28mn + 49n^{2})$

(A) The algebraic identity is: $x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)$.
We can rewrite the given expression as:
$27x^3 + y^3 + z^3 - 9xyz = (3x)^3 + (y)^3 + (z)^3 - 3(3x)(y)(z)$.
Comparing this with the identity,where $x$ is replaced by $3x$,$y$ by $y$,and $z$ by $z$,we get:
$= (3x + y + z)((3x)^2 + y^2 + z^2 - (3x)(y) - (y)(z) - (z)(3x))$.
Simplifying the terms inside the second bracket:
$= (3x + y + z)(9x^2 + y^2 + z^2 - 3xy - yz - 3zx)$.

(N/A) Start with the Right Hand Side ($R$.$H$.$S$.):
$\text{R.H.S.} = \frac{1}{2}(x+y+z)[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}]$
Expand the squares inside the bracket:
$= \frac{1}{2}(x+y+z)[(x^{2}-2xy+y^{2})+(y^{2}-2yz+z^{2})+(z^{2}-2zx+x^{2})]$
Combine like terms:
$= \frac{1}{2}(x+y+z)[2x^{2}+2y^{2}+2z^{2}-2xy-2yz-2zx]$
Factor out $2$ from the expression inside the square bracket:
$= \frac{1}{2}(x+y+z) \cdot 2[x^{2}+y^{2}+z^{2}-xy-yz-zx]$
$= (x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)$
Using the algebraic identity $x^{3}+y^{3}+z^{3}-3xyz = (x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)$,we get:
$= x^{3}+y^{3}+z^{3}-3xyz = \text{L.H.S.}$

(N/A) Given that $x+y+z=0.$
Therefore,$x+y=-z.$
Cubing both sides,we get $(x+y)^{3}=(-z)^{3}.$
Using the identity $(x+y)^{3} = x^{3}+y^{3}+3xy(x+y),$ we have:
$x^{3}+y^{3}+3xy(x+y) = -z^{3}.$
Since $x+y=-z,$ substitute this into the equation:
$x^{3}+y^{3}+3xy(-z) = -z^{3}.$
$x^{3}+y^{3}-3xyz = -z^{3}.$
Rearranging the terms,we get:
$x^{3}+y^{3}+z^{3} = 3xyz.$
Hence,if $x+y+z=0,$ then $x^{3}+y^{3}+z^{3}=3xyz.$

(C) Let $x = -12, y = 7,$ and $z = 5$.
Then,$x + y + z = -12 + 7 + 5 = 0$.
We know that if $x + y + z = 0$,then $x^{3} + y^{3} + z^{3} = 3xyz$.
Therefore,$(-12)^{3} + (7)^{3} + (5)^{3} = 3[(-12)(7)(5)]$,since $(-12) + 7 + 5 = 0$.
Calculating the product: $3[-420] = -1260$.
Thus,$(-12)^{3} + (7)^{3} + (5)^{3} = -1260$.

(D) Let $x = 28$,$y = -15$,and $z = -13$.
First,calculate the sum of $x, y,$ and $z$:
$x + y + z = 28 + (-15) + (-13) = 28 - 28 = 0$.
We know the algebraic identity: If $x + y + z = 0$,then $x^{3} + y^{3} + z^{3} = 3xyz$.
Substituting the values into the identity:
$(28)^{3} + (-15)^{3} + (-13)^{3} = 3(28)(-15)(-13)$.
Calculating the product:
$3 \times 28 = 84$
$(-15) \times (-13) = 195$
$84 \times 195 = 16380$.
Therefore,the value is $16380$.

(B) We know the algebraic identity: $x^{3}+y^{3}+z^{3}-3xyz = (x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)$.
If $x+y+z = 0$,then $x^{3}+y^{3}+z^{3} = 3xyz$.
We need to find the value of $48^{3}-30^{3}-18^{3}$,which can be written as $48^{3}+(-30)^{3}+(-18)^{3}$.
Let $x = 48$,$y = -30$,and $z = -18$.
Check the sum: $x+y+z = 48 + (-30) + (-18) = 48 - 48 = 0$.
Since the sum is $0$,we can use the identity $x^{3}+y^{3}+z^{3} = 3xyz$.
Therefore,$48^{3}+(-30)^{3}+(-18)^{3} = 3 \times 48 \times (-30) \times (-18)$.
Calculating the product: $3 \times 48 = 144$,and $(-30) \times (-18) = 540$.
$144 \times 540 = 77760$.

(D) We know the algebraic identity: $a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$.
If $a+b+c = 0$,then $a^{3}+b^{3}+c^{3} = 3abc$.
Let $a = (x-y)$,$b = (y-z)$,and $c = (z-x)$.
Now,calculate the sum $a+b+c = (x-y) + (y-z) + (z-x) = x - x + y - y + z - z = 0$.
Since the sum is $0$,we can apply the identity $a^{3}+b^{3}+c^{3} = 3abc$.
Therefore,$(x-y)^{3}+(y-z)^{3}+(z-x)^{3} = 3(x-y)(y-z)(z-x)$.

(N/A) Given the polynomial $p(y) = (y+2)(y-2)$.
Using the algebraic identity $(a+b)(a-b) = a^2 - b^2$,we can simplify the expression as $p(y) = y^2 - 4$.
Step $1$: Find $p(0)$.
$p(0) = (0)^2 - 4 = 0 - 4 = -4$.
Step $2$: Find $p(1)$.
$p(1) = (1)^2 - 4 = 1 - 4 = -3$.
Step $3$: Find $p(-2)$.
$p(-2) = (-2)^2 - 4 = 4 - 4 = 0$.

(D) To evaluate $103^{3}$,we can express it as $(100 + 3)^{3}$.
Using the algebraic identity $(a + b)^{3} = a^{3} + b^{3} + 3ab(a + b)$,where $a = 100$ and $b = 3$:
$(100 + 3)^{3} = (100)^{3} + (3)^{3} + 3(100)(3)(100 + 3)$
$= 1000000 + 27 + 900(103)$
$= 1000000 + 27 + 92700$
$= 1092727$

(A) We can write the expression as:
$101 \times 102 = (100 + 1)(100 + 2)$
Using the algebraic identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,where $x = 100$,$a = 1$,and $b = 2$:
$(100 + 1)(100 + 2) = (100)^2 + (1 + 2)(100) + (1)(2)$
$= 10000 + (3)(100) + 2$
$= 10000 + 300 + 2$
$= 10302$

(B) We can express $999$ as $(1000 - 1)$.
Using the algebraic identity $(a - b)^{2} = a^{2} - 2ab + b^{2}$,where $a = 1000$ and $b = 1$:
$(999)^{2} = (1000 - 1)^{2}$
$= (1000)^{2} - 2 \times (1000) \times (1) + (1)^{2}$
$= 1000000 - 2000 + 1$
$= 998000 + 1$
$= 998001$

(A) We have the expression $4 x^{2}+20 x+25$.
This can be written in the form $a^{2}+2 a b+b^{2}$ as follows:
$4 x^{2}+20 x+25 = (2 x)^{2}+2(2 x)(5)+(5)^{2}$
Using the algebraic identity $a^{2}+2 a b+b^{2}=(a+b)^{2}$,where $a = 2x$ and $b = 5$,we get:
$(2 x+5)^{2}$
Thus,the factors are $(2 x+5)(2 x+5)$.

(N/A) We have the expression $9 y^{2}-66 y z+121 z^{2}$.
This can be written in the form $a^{2}-2 a b+b^{2} = (a-b)^{2}$.
$9 y^{2}-66 y z+121 z^{2} = (3 y)^{2} - 2(3 y)(11 z) + (11 z)^{2}$.
Comparing this with the identity $(a-b)^{2} = a^{2}-2 a b+b^{2}$,where $a = 3 y$ and $b = 11 z$.
Therefore,$9 y^{2}-66 y z+121 z^{2} = (3 y - 11 z)^{2}$.
This can be written as $(3 y - 11 z)(3 y - 11 z)$.

We use the algebraic identity $a^{2}-b^{2}=(a+b)(a-b)$,where $a = \left(2x + \frac{1}{3}\right)$ and $b = \left(x - \frac{1}{2}\right)$.
Substituting these into the identity:
$= \left[\left(2x + \frac{1}{3}\right) + \left(x - \frac{1}{2}\right)\right] \left[\left(2x + \frac{1}{3}\right) - \left(x - \frac{1}{2}\right)\right]$
Simplify the terms inside the brackets:
$= \left(2x + x + \frac{1}{3} - \frac{1}{2}\right) \left(2x - x + \frac{1}{3} + \frac{1}{2}\right)$
$= \left(3x + \frac{2-3}{6}\right) \left(x + \frac{2+3}{6}\right)$
$= \left(3x - \frac{1}{6}\right) \left(x + \frac{5}{6}\right)$

(A) We have the expression $9x^{2}-12x+4$.
This can be written in the form $a^{2}-2ab+b^{2}$ where $a=3x$ and $b=2$.
$9x^{2}-12x+4 = (3x)^{2} - 2(3x)(2) + (2)^{2}$.
Using the algebraic identity $a^{2}-2ab+b^{2} = (a-b)^{2}$,we get:
$(3x-2)^{2} = (3x-2)(3x-2)$.

(A) To expand the expression $(4a - b + 2c)^2$,we use the algebraic identity:
$(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$
Here,$x = 4a$,$y = -b$,and $z = 2c$.
Substituting these values into the identity:
$(4a - b + 2c)^2 = (4a)^2 + (-b)^2 + (2c)^2 + 2(4a)(-b) + 2(-b)(2c) + 2(2c)(4a)$
Calculating each term:
$= 16a^2 + b^2 + 4c^2 - 8ab - 4bc + 16ac$

(A) To expand $(3a - 5b - c)^2$,we use the algebraic identity:
$(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$
Here,$x = 3a$,$y = -5b$,and $z = -c$.
Substituting these values into the identity:
$(3a - 5b - c)^2 = (3a)^2 + (-5b)^2 + (-c)^2 + 2(3a)(-5b) + 2(-5b)(-c) + 2(-c)(3a)$
Calculating each term:
$= 9a^2 + 25b^2 + c^2 - 30ab + 10bc - 6ca$

(A) We use the algebraic identity: $(a+b+c)^{2} = a^{2}+b^{2}+c^{2}+2ab+2bc+2ca$.
Here,$a = -x$,$b = 2y$,and $c = -3z$.
Substituting these values into the identity:
$(-x+2y-3z)^{2} = (-x)^{2} + (2y)^{2} + (-3z)^{2} + 2(-x)(2y) + 2(2y)(-3z) + 2(-3z)(-x)$
$= x^{2} + 4y^{2} + 9z^{2} - 4xy - 12yz + 6xz$.

(A) We have the expression: $9 x^{2}+4 y^{2}+16 z^{2}+12 x y-16 y z-24 x z$.
This can be written in the form $a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a = (a+b+c)^{2}$.
Here,$a = 3x$,$b = 2y$,and $c = -4z$.
Substituting these values:
$9 x^{2}+4 y^{2}+16 z^{2}+12 x y-16 y z-24 x z = (3 x)^{2}+(2 y)^{2}+(-4 z)^{2}+2(3 x)(2 y)+2(2 y)(-4 z)+2(-4 z)(3 x)$.
Using the identity $(a+b+c)^{2}$,we get:
$(3 x+2 y-4 z)^{2}$.
Thus,the factorised form is $(3 x+2 y-4 z)(3 x+2 y-4 z)$.

(N/A) The given expression is $25 x^{2}+16 y^{2}+4 z^{2}-40 x y+16 y z-20 x z$.
We use the algebraic identity: $(a+b+c)^{2} = a^{2}+b^{2}+c^{2}+2ab+2bc+2ca$.
Comparing the given expression with the identity,we can rewrite it as:
$=(-5 x)^{2}+(4 y)^{2}+(2 z)^{2}+2(-5 x)(4 y)+2(4 y)(2 z)+2(2 z)(-5 x)$.
Here,$a = -5x$,$b = 4y$,and $c = 2z$.
Therefore,the expression simplifies to $(-5 x+4 y+2 z)^{2}$ or $(5 x-4 y-2 z)^{2}$.

(N/A) The given expression is $16 x^{2}+4 y^{2}+9 z^{2}-16 x y-12 y z+24 x z$.
We can rewrite this expression using the identity $(a+b+c)^{2} = a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a$.
Here,we observe the terms:
$16 x^{2} = (4 x)^{2}$
$4 y^{2} = (-2 y)^{2}$
$9 z^{2} = (3 z)^{2}$
Now,checking the cross terms:
$2(4 x)(-2 y) = -16 x y$
$2(-2 y)(3 z) = -12 y z$
$2(3 z)(4 x) = 24 x z$
Thus,the expression becomes:
$(4 x)^{2}+(-2 y)^{2}+(3 z)^{2}+2(4 x)(-2 y)+2(-2 y)(3 z)+2(3 z)(4 x)$
Applying the identity,we get:
$(4 x-2 y+3 z)^{2}$
Therefore,the factorised form is $(4 x-2 y+3 z)(4 x-2 y+3 z)$.

(B) We know the algebraic identity:
$(a+b+c)^{2} = a^{2}+b^{2}+c^{2}+2(ab+bc+ca)$
Given that $a+b+c = 9$ and $ab+bc+ca = 26$.
Substituting these values into the identity:
$(9)^{2} = (a^{2}+b^{2}+c^{2}) + 2(26)$
$81 = (a^{2}+b^{2}+c^{2}) + 52$
Subtracting $52$ from both sides:
$a^{2}+b^{2}+c^{2} = 81 - 52$
$a^{2}+b^{2}+c^{2} = 29$

(A) To expand the expression $(3a - 2b)^3$,we use the algebraic identity:
$(x - y)^3 = x^3 - y^3 - 3xy(x - y)$
Here,$x = 3a$ and $y = 2b$.
Substituting these values into the identity:
$(3a - 2b)^3 = (3a)^3 - (2b)^3 - 3(3a)(2b)(3a - 2b)$
Calculate the cubes:
$(3a)^3 = 27a^3$
$(2b)^3 = 8b^3$
Now,simplify the expression:
$= 27a^3 - 8b^3 - 18ab(3a - 2b)$
Distribute $-18ab$ into the parentheses:
$= 27a^3 - 8b^3 - (18ab \times 3a) + (18ab \times 2b)$
$= 27a^3 - 8b^3 - 54a^2b + 36ab^2$

We use the algebraic identity: $(a+b)^{3} = a^{3} + b^{3} + 3ab(a+b)$.
Here,$a = \frac{1}{x}$ and $b = \frac{y}{3}$.
Substituting these values into the identity:
$\left(\frac{1}{x} + \frac{y}{3}\right)^{3} = \left(\frac{1}{x}\right)^{3} + \left(\frac{y}{3}\right)^{3} + 3 \left(\frac{1}{x}\right) \left(\frac{y}{3}\right) \left(\frac{1}{x} + \frac{y}{3}\right)$
$= \frac{1}{x^{3}} + \frac{y^{3}}{27} + \frac{y}{x} \left(\frac{1}{x} + \frac{y}{3}\right)$
$= \frac{1}{x^{3}} + \frac{y^{3}}{27} + \frac{y}{x^{2}} + \frac{y^{2}}{3x}$
Rearranging the terms,we get: $\frac{1}{x^{3}} + \frac{y}{x^{2}} + \frac{y^{2}}{3x} + \frac{y^{3}}{27}$.

(D) We use the algebraic identity $(a-b)^{3} = a^{3} - b^{3} - 3ab(a-b)$.
Here,$a = 4$ and $b = \frac{1}{3x}$.
Substituting these values into the identity:
$\left(4-\frac{1}{3 x}\right)^{3} = (4)^{3} - \left(\frac{1}{3 x}\right)^{3} - 3(4)\left(\frac{1}{3 x}\right)\left(4-\frac{1}{3 x}\right)$
$= 64 - \frac{1}{27 x^{3}} - \frac{12}{3 x}\left(4-\frac{1}{3 x}\right)$
$= 64 - \frac{1}{27 x^{3}} - \frac{4}{x}\left(4-\frac{1}{3 x}\right)$
$= 64 - \frac{1}{27 x^{3}} - \frac{16}{x} + \frac{4}{3 x^{2}}$

(A) The given expression is $8 p^{3}+\frac{12}{5} p^{2}+\frac{6}{25} p+\frac{1}{125}$.
We can rewrite this expression in the form of the algebraic identity $(a+b)^{3} = a^{3} + 3a^{2}b + 3ab^{2} + b^{3}$.
Here,$a = 2p$ and $b = \frac{1}{5}$.
Substituting these values into the identity:
$(2p)^{3} + 3(2p)^{2}(\frac{1}{5}) + 3(2p)(\frac{1}{5})^{2} + (\frac{1}{5})^{3}$
$= 8p^{3} + 3(4p^{2})(\frac{1}{5}) + 3(2p)(\frac{1}{25}) + \frac{1}{125}$
$= 8p^{3} + \frac{12}{5}p^{2} + \frac{6}{25}p + \frac{1}{125}$.
Thus,the expression is equivalent to $(2p + \frac{1}{5})^{3}$,which can be written as $(2p + \frac{1}{5})(2p + \frac{1}{5})(2p + \frac{1}{5})$.

(N/A) The given expression is of the form $(a+b)(a^2-ab+b^2)$,where $a = \frac{x}{2}$ and $b = 2y$.
Using the algebraic identity $(a+b)(a^2-ab+b^2) = a^3+b^3$:
$\left(\frac{x}{2}+2 y\right)\left\{\left(\frac{x}{2}\right)^{2}-\left(\frac{x}{2}\right)(2 y)+(2 y)^{2}\right\} = \left(\frac{x}{2}\right)^{3}+(2 y)^{3}$
$= \frac{x^3}{8} + 8y^3$

$(X^6-1)$ We use the algebraic identity $(a-b)(a^{2}+ab+b^{2}) = a^{3}-b^{3}$.
Given expression: $(x^{2}-1)(x^{4}+x^{2}+1)$
Rewrite the expression as: $(x^{2}-1)((x^{2})^{2} + (x^{2})(1) + (1)^{2})$
Here,$a = x^{2}$ and $b = 1$.
Applying the identity: $(x^{2})^{3} - (1)^{3}$
$= x^{6} - 1$

(A) We use the algebraic identity: $a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2})$.
Given expression: $a^{3} - 2\sqrt{2}b^{3}$.
Rewrite the expression as: $a^{3} - (\sqrt{2}b)^{3}$.
Applying the identity where $x = a$ and $y = \sqrt{2}b$:
$= (a - \sqrt{2}b)((a)^{2} + (a)(\sqrt{2}b) + (\sqrt{2}b)^{2})$.
$= (a - \sqrt{2}b)(a^{2} + \sqrt{2}ab + 2b^{2})$.

(D) We use the algebraic identity: $(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca) = a^{3}+b^{3}+c^{3}-3abc$.
Given expression: $(2x - y + 3z)((2x)^{2} + (-y)^{2} + (3z)^{2} - (2x)(-y) - (-y)(3z) - (3z)(2x))$.
Here,$a = 2x$,$b = -y$,and $c = 3z$.
Applying the identity:
$= (2x)^{3} + (-y)^{3} + (3z)^{3} - 3(2x)(-y)(3z)$
$= 8x^{3} - y^{3} + 27z^{3} - 3(2x)(-y)(3z)$
$= 8x^{3} - y^{3} + 27z^{3} + 18xyz$.

(A) We use the algebraic identity: $x^{3}+y^{3}+z^{3}-3xyz = (x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)$.
Given expression: $a^{3}-8b^{3}-64c^{3}-24abc$
This can be rewritten as: $a^{3}+(-2b)^{3}+(-4c)^{3}-3(a)(-2b)(-4c)$.
Here,$x=a$,$y=-2b$,and $z=-4c$.
Applying the identity:
$= (a-2b-4c)((a)^{2}+(-2b)^{2}+(-4c)^{2}-(a)(-2b)-(-2b)(-4c)-(-4c)(a))$
$= (a-2b-4c)(a^{2}+4b^{2}+16c^{2}+2ab-8bc+4ca)$.

(A) We use the algebraic identity: $x^{3} + y^{3} + z^{3} - 3xyz = (x + y + z)(x^{2} + y^{2} + z^{2} - xy - yz - zx)$.
Given expression: $2 \sqrt{2} a^{3} + 8 b^{3} - 27 c^{3} + 18 \sqrt{2} a b c$
This can be rewritten as: $(\sqrt{2} a)^{3} + (2 b)^{3} + (-3 c)^{3} - 3(\sqrt{2} a)(2 b)(-3 c)$.
Here,$x = \sqrt{2} a$,$y = 2 b$,and $z = -3 c$.
Applying the identity:
$= (\sqrt{2} a + 2 b - 3 c)((\sqrt{2} a)^{2} + (2 b)^{2} + (-3 c)^{2} - (\sqrt{2} a)(2 b) - (2 b)(-3 c) - (-3 c)(\sqrt{2} a))$
$= (\sqrt{2} a + 2 b - 3 c)(2 a^{2} + 4 b^{2} + 9 c^{2} - 2 \sqrt{2} a b + 6 b c + 3 \sqrt{2} c a)$.

(C) Let $a = \frac{1}{2}$,$b = \frac{1}{3}$,and $c = -\frac{5}{6}$.
First,calculate the sum $a + b + c$:
$a + b + c = \frac{1}{2} + \frac{1}{3} - \frac{5}{6} = \frac{3 + 2 - 5}{6} = \frac{0}{6} = 0$.
We know the algebraic identity: If $a + b + c = 0$,then $a^{3} + b^{3} + c^{3} = 3abc$.
Applying this identity to the given expression:
$\left(\frac{1}{2}\right)^{3} + \left(\frac{1}{3}\right)^{3} + \left(-\frac{5}{6}\right)^{3} = 3 \times \left(\frac{1}{2}\right) \times \left(\frac{1}{3}\right) \times \left(-\frac{5}{6}\right)$.
$= 3 \times \frac{1}{6} \times \left(-\frac{5}{6}\right) = \frac{1}{2} \times \left(-\frac{5}{6}\right) = -\frac{5}{12}$.

(D) We have the expression: $(0.2)^{3} - (0.3)^{3} + (0.1)^{3}$.
This can be rewritten as: $(0.2)^{3} + (-0.3)^{3} + (0.1)^{3}$.
Let $a = 0.2$,$b = -0.3$,and $c = 0.1$.
Now,calculate the sum $a + b + c$:
$a + b + c = 0.2 + (-0.3) + 0.1 = 0.3 - 0.3 = 0$.
We know the algebraic identity: If $a + b + c = 0$,then $a^{3} + b^{3} + c^{3} = 3abc$.
Substituting the values of $a$,$b$,and $c$:
$a^{3} + b^{3} + c^{3} = 3(0.2)(-0.3)(0.1)$.
Calculating the product:
$3 \times 0.2 = 0.6$
$0.6 \times (-0.3) = -0.18$
$-0.18 \times 0.1 = -0.018$.
Therefore,$(0.2)^{3} - (0.3)^{3} + (0.1)^{3} = -0.018$.

(D) Let $a = x-2y$,$b = 2y-3z$,and $c = 3z-x$.
Then,$a+b+c = (x-2y) + (2y-3z) + (3z-x) = 0$.
We know that if $a+b+c = 0$,then $a^3 + b^3 + c^3 = 3abc$.
Substituting the values of $a, b,$ and $c$ back into the identity,we get:
$(x-2y)^3 + (2y-3z)^3 + (3z-x)^3 = 3(x-2y)(2y-3z)(3z-x)$.

(B) Given expression: $x^{3}+y^{3}-12xy+64$.
We know the algebraic identity: $a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$.
Here,let $a=x$,$b=y$,and $c=4$.
Then the expression becomes $x^{3}+y^{3}+4^{3}-3(x)(y)(4) = x^{3}+y^{3}+64-12xy$.
Substituting $a+b+c = x+y+4$ into the identity:
Since $x+y = -4$,we have $x+y+4 = -4+4 = 0$.
Therefore,$x^{3}+y^{3}-12xy+64 = (0)(x^{2}+y^{2}+16-xy-4y-4x) = 0$.

(C) Given expression is $x^{3}-8 y^{3}-36 x y-216$.
We can rewrite this as $x^{3} + (-2y)^{3} + (-6)^{3} - 3(x)(-2y)(-6)$.
This is in the form of the algebraic identity $a^{3} + b^{3} + c^{3} - 3abc = (a+b+c)(a^{2} + b^{2} + c^{2} - ab - bc - ca)$,where $a=x$,$b=-2y$,and $c=-6$.
Substituting these values,we get: $(x - 2y - 6)(x^{2} + 4y^{2} + 36 + 2xy - 12y + 6x)$.
Since we are given $x = 2y + 6$,it follows that $x - 2y - 6 = 0$.
Therefore,the entire expression becomes $0 \times (x^{2} + 4y^{2} + 36 + 2xy - 12y + 6x) = 0$.

(A) We use the algebraic identity: $x^{3}+y^{3} = (x+y)^{3} - 3xy(x+y)$.
Given $x+y = 12$ and $xy = 27$.
Substituting these values into the identity:
$x^{3}+y^{3} = (12)^{3} - 3(27)(12)$.
Factor out $12$ to simplify the calculation:
$x^{3}+y^{3} = 12(12^{2} - 3 \times 27)$.
$x^{3}+y^{3} = 12(144 - 81)$.
$x^{3}+y^{3} = 12(63)$.
$x^{3}+y^{3} = 756$.

(A) We use the algebraic identity $a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})$.
Let $a = (2x - 5y)$ and $b = (2x + 5y)$.
Then,$a - b = (2x - 5y) - (2x + 5y) = 2x - 5y - 2x - 5y = -10y$.
Now,calculate $a^{2} + ab + b^{2}$:
$a^{2} = (2x - 5y)^{2} = 4x^{2} + 25y^{2} - 20xy$
$b^{2} = (2x + 5y)^{2} = 4x^{2} + 25y^{2} + 20xy$
$ab = (2x - 5y)(2x + 5y) = 4x^{2} - 25y^{2}$
Summing these: $(4x^{2} + 25y^{2} - 20xy) + (4x^{2} - 25y^{2}) + (4x^{2} + 25y^{2} + 20xy) = 12x^{2} + 25y^{2}$.
Finally,multiply $(a - b)$ and $(a^{2} + ab + b^{2})$:
$(-10y)(12x^{2} + 25y^{2}) = -120x^{2}y - 250y^{3}$.

(D) We have the expression: $(-z+x-2 y)(x^{2}+4 y^{2}+z^{2}+2 x y+x z-2 y z)$.
Rearranging the terms,we get: $(x-2 y-z)(x^{2}+(-2 y)^{2}+(-z)^{2}-(x)(-2 y)-(-2 y)(-z)-(x)(-z))$.
This expression is in the form $(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$,where $a=x$,$b=-2 y$,and $c=-z$.
Using the algebraic identity $(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca) = a^{3}+b^{3}+c^{3}-3abc$,we substitute the values:
$= (x)^{3} + (-2 y)^{3} + (-z)^{3} - 3(x)(-2 y)(-z)$.
$= x^{3} - 8 y^{3} - z^{3} - 6 x y z$.

(N/A) Given that $a, b, c$ are non-zero and $a+b+c=0$.
We know the algebraic identity: $a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$.
Since $a+b+c=0$,the right side becomes $0$,so $a^{3}+b^{3}+c^{3}-3abc = 0$,which implies $a^{3}+b^{3}+c^{3}=3abc$.
Now,consider the expression $\frac{a^{2}}{bc} + \frac{b^{2}}{ca} + \frac{c^{2}}{ab}$.
Taking the least common multiple $(LCM)$ of the denominators $bc, ca, ab$,which is $abc$,we get:
$\frac{a^{2}(a) + b^{2}(b) + c^{2}(c)}{abc} = \frac{a^{3}+b^{3}+c^{3}}{abc}$.
Substituting $a^{3}+b^{3}+c^{3} = 3abc$ into the expression:
$\frac{3abc}{abc} = 3$.
Thus,the expression equals $3$.

(N/A) We know the algebraic identity:
$a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
$= (a+b+c)[a^2+b^2+c^2-(ab+bc+ca)]$
Given $a+b+c=5$ and $ab+bc+ca=10$,we substitute these values:
$= 5[a^2+b^2+c^2-10]$
Now,we find $a^2+b^2+c^2$ using the identity $(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca)$:
$(5)^2 = a^2+b^2+c^2+2(10)$
$25 = a^2+b^2+c^2+20$
$a^2+b^2+c^2 = 25-20 = 5$
Substituting this value back into the expression:
$a^3+b^3+c^3-3abc = 5(5-10) = 5(-5) = -25$.
Hence,it is proved.

(N/A) We know the expansion of $(a+b+c)^3$ is given by:
$(a+b+c)^3 = a^3 + b^3 + c^3 + 3(a+b)(b+c)(c+a)$
Alternatively,we can expand it step-by-step:
$(a+b+c)^3 = [(a+b)+c]^3$
$= (a+b)^3 + 3(a+b)^2c + 3(a+b)c^2 + c^3$
$= (a^3 + 3a^2b + 3ab^2 + b^3) + 3(a^2 + 2ab + b^2)c + 3ac^2 + 3bc^2 + c^3$
$= a^3 + b^3 + c^3 + 3a^2b + 3ab^2 + 3a^2c + 6abc + 3b^2c + 3ac^2 + 3bc^2$
Now,subtract $(a^3 + b^3 + c^3)$ from both sides:
$(a+b+c)^3 - a^3 - b^3 - c^3 = 3a^2b + 3ab^2 + 3a^2c + 6abc + 3b^2c + 3ac^2 + 3bc^2$
Factor out $3$ from the right side:
$= 3[a^2b + ab^2 + a^2c + 2abc + b^2c + ac^2 + bc^2]$
$= 3[ab(a+b) + ac(a+c) + bc(b+c) + 2abc]$
$= 3[ab(a+b) + ac(a+c) + bc(b+c) + abc + abc]$
$= 3[ab(a+b) + abc + ac(a+c) + ac^2 + bc(b+c) + abc]$
$= 3[ab(a+b+c) + ac(a+c+b) + bc(b+c+a)]$
$= 3(a+b+c)(ab + ac + bc)$
Wait,the standard identity is $(a+b+c)^3 - a^3 - b^3 - c^3 = 3(a+b)(b+c)(c+a)$.
Let's verify: $3(a+b)(b+c)(c+a) = 3(ab + ac + b^2 + bc)(c+a)$
$= 3(abc + a^2b + ac^2 + a^2c + b^2c + ab^2 + bc^2 + abc)$
$= 3(a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2 + 2abc)$.
This matches the expansion derived above.
Hence,$(a+b+c)^3 - a^3 - b^3 - c^3 = 3(a+b)(b+c)(c+a)$ is proved.

(A) To expand the expression $(x+3)(x+8)$,we use the algebraic identity $(x+a)(x+b) = x^2 + (a+b)x + ab$.
Here,$a = 3$ and $b = 8$.
Substituting these values into the identity:
$(x+3)(x+8) = x^2 + (3+8)x + (3)(8)$
$= x^2 + 11x + 24$.

(A) To expand the expression $(2 x-3)(2 x+5)$,we use the algebraic identity $(x+a)(x+b) = x^2 + (a+b)x + ab$.
Here,let $X = 2x$,$a = -3$,and $b = 5$.
Substituting these values into the identity:
$(2x - 3)(2x + 5) = (2x)^2 + (-3 + 5)(2x) + (-3)(5)$
$= 4x^2 + (2)(2x) - 15$
$= 4x^2 + 4x - 15$.

(D) To expand the expression $(3x - 2)(3x - 6)$,we use the distributive property or the algebraic identity $(x + a)(x + b) = x^2 + (a + b)x + ab$.
Given expression: $(3x - 2)(3x - 6)$
Let $u = 3x$. Then the expression becomes $(u - 2)(u - 6)$.
Expanding this: $u^2 + (-2 - 6)u + (-2)(-6)$
$= u^2 - 8u + 12$
Now,substitute $u = 3x$ back into the expression:
$= (3x)^2 - 8(3x) + 12$
$= 9x^2 - 24x + 12$

(A) To expand the expression $(x+2t)(x-5t)$,we use the algebraic identity $(x+a)(x+b) = x^2 + (a+b)x + ab$.
Here,$a = 2t$ and $b = -5t$.
Substituting these values into the identity:
$(x+2t)(x-5t) = x^2 + (2t - 5t)x + (2t)(-5t)$
$= x^2 + (-3t)x - 10t^2$
$= x^2 - 3xt - 10t^2$.