$bold{a)};;;$ $2x(3x-5x^2+5y)$

$=2x(3x)+2x(-5x^2)+2x(4y)$

$$
= 6x-10x^3+8xy
$$

Apply distributive property,

$a(b+c)=ab+ac$

$bold{b)};;;$ $(3x-4)(2x+5)$

$=(3x-4)(2x)+(3x-4)(5)$

$=[(3x)(2x)-4(2x)]+[(3x)(5)-4(5)]$

$=(6x^2-8x)+(15x-20)$

$=6x^2+7x-20$

Apply distributive property,

$$
(a+b)(c+d)=(a+b)(c)+(a+b)(d)
$$

$bold{c);;;}$ $(x+4)^2$

$=(x)^2+2(x)(4)+(4)^2$

$=x^2+8x+16$

You may use special product formula:

$(a+b)^2=a^2+2ab+b^2$

$bold{d);;;}$ $(x+1)(x^2+2x-3)$

$=(x+1)(x^2)+(x+1)(2x)+(x+1)(-3)$

$=(x^3+x^2)+(2x^2+2x)+(-3x-3)$

$=x^3+(x^2+2x^2)+(2x-3x)-3$

$$
=x^3+3x^2-x-3
$$

Apply distributive property,

$$
(a+b)(c+d)=(a+b)(c)+(a+b)(d)
$$

a) $6x^2-10x^3+8xy$

b) $6x^2+7x-20$

c) $x^2+8x+16$

d) $x^3+3x^2-x-3$

Let $x=1$ and simplify the expressions

$(3(1)+2)^2neq9(1)^2+4$

$5^2neq9+4$

$$
25neq13
$$

$(3x+2)(3x+2)$

$3x(3x)+2(3x)+2(3x)+2(2)$

$9x^2+6x+6x+4$

$$
9x^2+12x+4
$$

a) No! When $x=1$, left side is 25 while right side is 13.

b) $9x^2+12x+4$

Remember the distributive property: $(a+b)(c+d)=(a+b)(c)+(a+b)(d)$. Here, we shall solve it by multiplying from left to right and from right to left. We expect to get the same answer.

$$
begin{align*} bold{a);;;} (2x+4)(3x^2+6x-5)&=(2x+4)(3x^2)+(2x+4)(6x)+(2x+4)(-5)\\
&=[(2x)(3x^2)+4(3x^2)]+[(2x)(6x)+4(6x)]\\
&quad+[(2x)(-5)+4(-5)]\\
&=(6x^3+12x^2)+(12x^2+24x)+(-10x-20)\\
&=(6x^3)+(12x^2+12x^2)+(-10x+24x)-20\\
&=6x^3+24x^2+14x-20end{align*}
$$

$$
begin{align*} bold{b);;;} (2x+4)(3x^2+6x-5)&=(2x)(3x^2+6x-5)+4(3x^2+6x-5)\\
&=[(2x)(3x^2)+(2x)(6x)+(2x)(-5)]\\
&quad+[4(3x^2)+4(6x)+4(-5)]\\
&=(6x^3+12x^2-10x)+(12x^2+24x-20)\\
&=(6x^3)+(12x^2+12x^2)+(-10x+24x)-20\\
&=6x^3+24x^2+14x-20\\
end{align*}
$$

a) $6x^3+24x^2+14x-20$

b) same as (a)

a)

$$
begin{align*}
5x(5x^{2}+3x-4)&=25x^{3}+15x^{2}-20x tag{text{use distributive property}}\
end{align*}
$$

b)

$$
begin{align*}
(x-6)(2x+5)&=2x^{2}+5x-12x-30 tag{text{use distributive property}}\
&=2x^{2}-7x-30 tag{text{simplify}}\
end{align*}
$$

c)

$$
begin{align*}
(x+3)(x-3)+(5x-6)(3x-7)&=\
&=x^{2}-3x+3x-9+15x^{2}-35x-18x+42 tag{text{multiply binomials and remove parentheses}}\
&=x^{2}+15x^{2}-3x+3x-35x-18x-9+52 tag{text{combine like terms}}\
&=16x^{2}-53x+43\
end{align*}
$$

d)

$$
begin{align*}
4(n-4)(3+n)-3(n-5)(n+8)&=\
&=4(3n+n^{2}-12-4n)-3(n^{2}+8n-5n-40) tag{text{multiply parehtheses}}\
&=4(n^{2}-n-12)-3(n^{2}+3n-40) tag{text{simplify expression in brackets }}\
&=4n^{2}-4n-48-3n^{2}-9n+120 tag{text{distributive property}}\
&=4n^{2}-3n^{2}-4n-9n-48+120 tag{text{combine like tems}}\
&=n^{2}-13n+72\
end{align*}
$$

e)

$$
begin{align*}
3(2x-1)^{2}-5(4x+1)^{2}&=\
&=3((2x)^{2}-2cdot2xcdot1+1)-5((4x)^{2}+2cdot4xcdot1+1)\
&=3(4x^{2}-4x+1)-5(16x^{2}+8x+1)\
&=12x^{2}-12x+3-80x^{2}-40x-5\
&=-68x^{2}-52x-2\
end{align*}
$$

d)

$$
begin{align*}
2(3a+4)(a-6)-(3-a)^{2}+4(5-a)&=\
&=2(3a^{2}-18a+4a-24)-(3^{2}-2cdot3cdot a+a^{2})+20-4a\
&=6a^{2}-36a+8a-48-9+6a-a^{2}+20-4a\
&=5a^{2}-26a-37\
end{align*}
$$

a) $25x^3+15x^2-20x$

b) $2x^2-7x-30$

c) $16x^2-53x+33$

d) $n^2-3n+72$

e) $-68x^2-52x-2$

f) $5a^2-26a-37$

a) Using the product of the sum and difference of like terms which is given by $(a+b)(a-b)=a^2-b^2,$ the given expression, $4x(x+5)(x-5)
,$ is equivalent

$$
begin{align*}
&
4x[(x)^2-(5)^2]
\&=
4x[x^2-25]
.end{align*}
$$

Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the
expression
above is equivalent to

$$
begin{align*}
&
4x(x^2)+4x(-25)
\&=
4x^3-100x
.end{align*}
$$

b) Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the given expression, $-2a(a+4)^2
,$ is equivalent to

$$
begin{align*}
&
-2a[(a)^2+2(a)(4)+(4)^2]
\&=
-2a[a^2+8a+16]
.end{align*}
$$

Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the
expression,
above is equivalent to

$$
begin{align*}
&
-2a(a^2)-2a(8a)-2a(16)
\&=
-2a^3-16a^2-32a
.end{align*}
$$

c) Using the product of the sum and difference of like terms which is given by $(a+b)(a-b)=a^2-b^2,$ the given expression, $(x+2)(x-5)(x-2)
,$ is equivalent

$$
begin{align*}
&
[(x+2)(x-2)](x-5)
\&=
[(x)^2-(2)^2](x-5)
\&=
(x^2-4)(x-5)
.end{align*}
$$

Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the expression above is equivalent to

$$
begin{align*}
&
x^2(x)+x^2(-5)-4(x)-4(-5)
\&=
x^3-5x^2-4x+20
.end{align*}
$$

d) Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the given expression, $(2x+1)(3x-5)(4-x)
,$ is equivalent to

$$
begin{align*}
&
[(2x+1)(3x-5)](4-x)
\&=
[2x(3x)+2x(-5)+1(3x)+1(-5)](4-x)
\&=
[6x^2-10x+3x-5](4-x)
\&=
(6x^2-7x-5)(4-x)
.end{align*}
$$

Using the Distributive Property, the expression above is equivalent to

$$
begin{align*}
&
6x^2(4)+6x^2(-x)-7x(4)-7x(-x)-5(4)-5(-x)
\&=
24x^2-6x^3-28x+7x^2-20+5x
\&=
-6x^3+(24x^2+7x^2)+(-28x+5x)-20
\&=
-6x^3+31x^2-23x-20
.end{align*}
$$

e) Using the Cube of a Binomial which is given by $(a+b)^3=a^3+3a^2b+3ab^2+b^3$ or $(a-b)^3=a^3-3a^2b+3ab^2-b^3,$ the given expression, $(9a-5)^3
,$ is equivalent to

$$
begin{align*}
&
(9a)^3-3(9a)^2(5)+3(9a)(5)^2-(5)^3
\&=
729a^3-1215a^2+675a-125
.end{align*}
$$

f) By the Commutative Property of Addition, the given expression, $(a-b+c-d)(a+b-c-d)
,$ is equivalent to

$$
begin{align*}
&
(a-d-b+c)(a-d+b-c)
.end{align*}
$$

By regrouping, the expression above is equivalent to

$$
begin{align*}
[(a-d)-(b-c)][(a-d)+(b-c)]
.end{align*}
$$

Using the product of the sum and difference of like terms which is given by $(a+b)(a-b)=a^2-b^2,$ the expression above is equivalent to

$$
begin{align*}
&
(a-d)^2-(b-c)^2
.end{align*}
$$

Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the expression above is equivalent to

$$
begin{align*}
&
(a^2-2ad+d^2)-(b^2-2bc+c^2)
\&=
a^2-2ad+d^2-b^2+2bc-c^2
\&=
a^2-b^2-c^2+d^2-2ad+2bc
.end{align*}
$$

a) $4x^3-100x$

b) $-2a^3-16a^2-32a$

c) $x^3-5x^2-4x+20$

d) $-6x^3+31x^2-23x-20$

e) $729a^3-1215a^2+675a-125$

f) $a^2-b^2-c^2+d^2-2ad+2bc$

a) Using the Distributive Property, the given expression, $(3x-2)(2x-1)
,$ is equivalent to

$$
begin{align*}
&
3x(2x-1)-2(2x-1)
.end{align*}
$$

Hence, the given expressions are equivalent.

b) Using the Distributive Property, the given expression, $(x-4)(2x^2+5x-6)
,$ is equivalent to

$$
begin{align*}
&
(2x^2+5x-6)(x-4)
\&=
2x^2(x-4)+5x(x-4)-6(x-4)
.end{align*}
$$

Hence, the given expressions are equivalent.

c) The given expression, $(x+2)(3x-1)-(1-2x)^2
,$ is equivalent to

$$
begin{align*}
&
x(3x)+x(-1)+2(3x)+2(-1)-(1-2x)^2
\&=
3x^2-x+6x-2-(1-2x)^2
\&=
3x^2+5x-2-(1-2x)^2
\&=
3x^2+5x-2-[(1)^2-2(1)(2x)+(2x)^2]
\&=
3x^2+5x-2-[1-4x+4x^2]
\&=
3x^2+5x-2-1+4x-4x^2
\&=
(3x^2-4x^2)+(5x+4x)+(-2-1)
\&=
-x^2+9x-3
.end{align*}
$$

The expression above is NOT equivalent to the other given expression, $x^2+9x-3.$

d) The given expression, $2(x-3)(2x^2-4x+5)
,$ is equivalent to

$$
begin{align*}
&
2[(x(2x^2)+x(-4x)+x(5)-3(2x^2)-3(-4x)-3(5)]
\&=
2[2x^3-4x^2+5x-6x^2+12x-15]
\&=
2[2x^3+(-4x^2-6x^2)+(5x+12x)-15]
\&=
2[2x^3-10x^2+17x-15]
\&=
2(2x^3)+2(-10x^2)+2(17x)+2(-15)
\&=
4x^3-20x^2+34x-30
.end{align*}
$$

This expression is equivalent to the other given expression.

e) The given expression, $(4x+y-3)^2
,$ is equivalent to

$$
begin{align*}
&
(4x)^2+(y)^2+(-3)^2+2(4x)(y)+2(4x)(-3)+2(y)(-3)
\&=
16x^2+y^2+9+8xy-24x-6y
.end{align*}
$$

This expression is NOT equivalent to the other given expression, $16x^2-8xy+24x+y^2-6y+9.$

f) The given expression, $3(y-2x)^3
,$ is equivalent to

$$
begin{align*}
&
3[(y)^3-3(y)^2(2x)+3(y)(2x)^2-(2x)^3]
\&=
3[y^3-3(y^2)(2x)+3(y)(4x^2)-(8x^3)]
\&=
3[y^3-6xy^2+12x^2y-8x^3]
\&=
3(y^3)+3(-6xy^2)+3(12x^2y)+3(-8x^3)
\&=
3y^3-18xy^2+36x^2y-24x^3
\&=
-24x^3+36x^2y-18xy^2+3y^3
.end{align*}
$$

This expression is equivalent to the other given expression.

If both sides of the equation is identical after simplification, then it is true for all real numbers. In this case, apply distributive property: $(a+b)(c+d)=(a+b)(c)+(a+b)(d)$

$$
begin{align*} (x-1)(x^4+x^3+x^2+x+1) &= (x-1)(x^4)+(x-1)(x^3)+(x-1)(x^2)\
&quad+(x-1)(x)+(x-1)\
&=(x^5-x^4)+(x^4-x^3)+(x^3-x^2)\
&quad+(x^2-x)+(x-1)\
&=x^5+(-x^4+x^4)+(-x^3+x^3)\
&quad+(-x^2+x^2)+(-x+x)-1\
&=x^5-1\
end{align*}
$$

Therefore, the equation $(x-1)(x^4+x^3+x^2+x+1) =x^5-1$ is true for all real numbers since the left side and right side expressions are equivalent.

a) Using $(ab)c$, the given expression, $19(5x+7)(3x-2)
,$ is equivalent to

$$
begin{align*}
&
[19(5x+7)](3x-2)
\&=
[95x+133](3x-2)
\&=
95x(3x)+95x(-2)+133(3x)+133(-2)
\&=
285x^2-190x+399x-266
\&=
285x^2+209-266
.end{align*}
$$

Using $a(bc)$, the given expression, $19(5x+7)(3x-2)
,$ is equivalent to

$$
begin{align*}
&
19[(5x+7)(3x-2)]
\&=
19[5x(3x)+5x(-2)+7(3x)+7(-2)]
\&=
19[15x^2-10x+21x-14]
\&=
19[15x^2+11x-14]
\&=
19(15x^2)+19(11x)+19(-14)
\&=
285x^2+209x-266
.end{align*}
$$

Both solutions above ended with the same answer. This verifies the Associative Property of Multiplication.

b) The second method is easier since its solution involves manipulating smaller numbers.

a) Both methods lead to $285x^2+209x-266$

b) The second method is easier as it involves smaller numbers.

a) Using $SA=2pi r^2+2pi rh
,$ with a radius of $(2x+1)$ and a height of $2x-1,$ then

$$
begin{align*}
SA&=2pi (2x+1)^2+2pi (2x+1)(2x-1)
\
SA&=2pi (2x+1)[(2x+1)+(2x-1)]
&text{ (factor the $GCF$)}
\
SA&=2pi (2x+1)(2x+1+2x-1)
\
SA&=2pi (2x+1)(4x)
\
SA&=8pi x(2x+1)
\
SA&=16pi x^2+8pi x
.end{align*}
$$

b) Using $V=pi r^2h
,$ with a radius of $(2x+1)$ and a height of $(2x-1),$ then

$$
begin{align*}
V&=pi (2x+1)^2(2x-1)
\
V&=pi [(2x)^2+2(2x)(1)+(1)^2](2x-1)
\
V&=pi (4x^2+4x+1)(2x-1)
\
V&=pi [4x^2(2x)+4x^2(-1)+4x(2x)+4x(-1)+1(2x)+1(-1)]
\
V&=pi [8x^3-4x^2+8x^2-4x+2x-1)
\
V&=pi [8x^3+4x^2-2x-1)
\
V&=8x^3pi+4x^2pi-2xpi-pi
.end{align*}
$$

a) $SA=16pi x^2+8pi x$

b) $V=8x^3pi+4x^2pi-2xpi-pi$

a) Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ then

$$
begin{align*}
(x-3)^2&Rightarrow(x)^2-2(x)(3)+(3)^2
\&=
x^2-6x+9
\\&text{and}\\
(3-x)^2&Rightarrow
(3)^2-2(3)(x)+(x)^2
\&=
9-6x+x^2
.end{align*}
$$

Hence, $(x-3)^2$ and $(3-x)^2$ are equivalent.

b) Using the cube of a binomial which is given by $(a+b)^3=a^3+3a^2b+3ab^2+b^3$ or $(a-b)^3=a^3-3a^2b+3ab^2-b^3,$ then

$$
begin{align*}
(x-3)^3&Rightarrow
x^3-3(x)^2(3)+3(x)(3)^2-(3)^3
\&=
x^3-3(x^2)(3)+3(x)(9)-(27)
\&=
x^3-9x^2+27x-27
\\&text{and}\\
(3-x)^3&Rightarrow
(3)^3-3(3)^2(x)+3(3)(x)^2-(x)^3
\&=
(27)-3(9)(x)+3(3)(x^2)-(x^3)
\&=
27-27x+9x^2-x^3
.end{align*}
$$

Hence, $(x-3)^3$ and $(3-x)^3$ are NOT equivalent.

a) Using the square of a multinomial which is given by $(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2b
,$ the given expression, $(x^2+2x-1)^2
,$ is equivalent to

$$
begin{align*}
&
(x)^2+(2x)^2+(-1)^2+2(x^2)(2x)+2(x^2)(-1)+2(2x)(-1)
\&=
x^2+4x^2+1+4x^3-2x^2-4x
.end{align*}
$$

b) Using the cube of a binomial which is given by

$$
begin{align*}
(a+b)^3&=a^3+3a^2b+3ab^2+b^3
\&text{or}\
(a-b)^3&=a^3-3a^2b+3ab^2-b^3
,end{align*}
$$
the given expression, $(2-a)^3
,$ is equivalent to

$$
begin{align*}
&
(2)^3-3(2)^2(a)+3(2)(a)^2-(a)^3
\&=
(8)-3(4)(a)+3(2)(a^2)-(a^3)
\&=
8-12a+6a^2-a^3
.end{align*}
$$

c) Grouping the terms, the given expression, $(x^3+x^2+x+1)(x^3-x^2-x-1)
,$ is equivalent to

$$
begin{align*}
&
[x^3+(x^2+x+1)][x^3-(x^2+x+1)]
.end{align*}
$$

Using the product of the sum and difference of like terms which is given by $(a+b)(a-b)=a^2-b^2,$ the expression above is equivalent to

$$
begin{align*}
&
(x^3)^2-(x^2+x+1)^2
\&=
x^6-(x^2+x+1)^2
.end{align*}
$$

Using the square of a multinomial which is given by $(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2b
,$ the expression above is equivalent to

$$
begin{align*}
&
x^6-[(x^2)^2+(x)^2+(1)^2+2(x^2)(x)+2(x^2)(1)+2(x)(1)]
\&=
x^6-[x^4+x^2+1+2x^3+2x^2+2x]
\&=
x^6-[x^4+2x^3+(x^2+2x^2)+2x+1]
\&=
x^6-[x^4+2x^3+3x^2+2x+1]
\&=
x^6-x^4-2x^3-3x^2-2x-1
.end{align*}
$$

d) Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the given expression, $2(x+1)^2-3(2x-1)(3x-5)
,$ is equivalent to

$$
begin{align*}
&
2[(x)^2+2(x)(1)+(1)^2]-3(2x-1)(3x-5)
\&=
2[x^2+2x+1]-3(2x-1)(3x-5)
.end{align*}
$$

Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the expression above is equivalent to

$$
begin{align*}
&
2[x^2+2x+1]-3[2x(3x)+2x(-5)-1(3x)-1(-5)]
\&=
2[x^2+2x+1]-3[6x^2-10x-3x+5]
\&=
2[x^2+2x+1]-3[6x^2-13x+5]
.end{align*}
$$

Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the
expression,
above is equivalent to

$$
begin{align*}
&
2x^2+4x+2-18x^2+39x-15
\&=
(2x^2-18x^2)+(4x+39x)+(2-15)
\&=
-16x^2+43x-13
.end{align*}
$$

a) $x^2+4x^2+1+4x^3-2x^2-4x$

b) $8-12a+6a^2-a^3$

c) $x^6-x^4-2x^3-3x^2-2x-1$

d) $-16x^2+43x-13$

For a right triangle with legs $x$ and $y$, the area $A$ is given by

$$
begin{equation*}A=frac{1}{2}xyend{equation*}
$$

If one side is doubled and the other side is halved, the equation becomes

$$
begin{equation*} A_2=frac{1}{2}(2x)left(dfrac{y}{2}right)=dfrac{1}{2}xy end{equation*}
$$

which is identical to the original equation. The change in area $(A-A_2)$ is

$$
begin{equation*} A-A_2 = dfrac{1}{2}xy-dfrac{1}{2}xy=0end{equation*}
$$

This means there will be no change in area if the one side is doubled and the other side is halved.

$0$ , that is, no change in area

The kinetic energy of the object is given by

$E=dfrac{1}{2}mv^2$

$bold{a);;;}$ if mass is increased by $x$, replace $m$ by $(m+x)$

$E=dfrac{1}{2}(m+x)(v^2)$

$E=dfrac{1}{2}mv^2+dfrac{1}{2}xv^2$

$bold{b);;;}$ if speed is increased by $y$, replace $v$ with $(v+y)$

$E=dfrac{1}{2}m(v+y)^2$

$E=dfrac{1}{2}m(v^2+2vy+y^2)$

$$
E=dfrac{1}{2}mv^2+mvy+dfrac{1}{2}my^2
$$

a) $E=dfrac{1}{2}mv^2+dfrac{1}{2}xv^2$

b) $E=dfrac{1}{2}mv^2+mvy+dfrac{1}{2}my^2$

$bold{a);;;}$ Investigate by observing the number of terms for different cases for $f(x)$ and $g(x)$:

(i) $f(x)$ has 2 terms and $g(x)$ has 1 term:

Example: $(a+b)(c)=ab+acimplies 2$ terms

(ii) If $f(x)$ has 2 terms and $g(x)$ has 2 terms

Example: $(a+b)(c+d)=ac+ad+bc+bdimplies$ 4 terms

(iii) $f(x)$ has 3 terms and $g(x)$ has 2 terms

Example: $(a+b+c)(d+e)=a(d+e)+b(d+e)+c(d+e)=ad+ae+bd+be+cd+ceimplies 6$ terms

It appears that if $f(x)$ has $n$ terms and $g(x)$ has $m$ terms, then the product contains $ntimes m$ terms

$bold{b);;;}$ If we have more than 2 factors. For example, if we have 3 factors with 2 terms each, if the pattern is still valid, we would expect that the number of terms would be $2times 2 times 2=8$

$$
begin{align*} (a+b)(c+d)(e+f)&=[(a+b)(c+d)](e+f)\
&=(ac+ad+bc+bd)(e+f)\
&=ac(e+f)+ad(e+f)+bc(e+f)+bd(e+f)\
&=ace+acf+ade+adf+bce+bcf+bde+bdf\
end{align*}
$$

Indeed, we have 8 terms. You could investigate further examples and it would show that the number of terms in the product of polynomials (without combining like terms) is equal to the product of the number of terms in each polynomial.

a) $f(x)$ has 2 terms and $g(x)$ has 3 terms

$(a+b+c)(d+e)=ad+ae+bd+be+cd+ceimplies 2times 3= 6$ terms

b) The number of terms in the product of polynomials (without combining like terms) is equal to the product of the number of terms in each polynomial.

a.i) The cubes that will have $3$ faces that are colored red are the cubes at the corner. There are a total of $8$ cubes at the corners.

a.ii) The cubes that will have $2$ faces that are colored red are the ones at the perimeter of the cube (except the corner cubes). There are a total of $12$ such cubes.

a.iii) The cubes that will have $1$ face that is colored red are the middle cubes. There are $6$ such cubes.

a.iv) There is 1 cube (innermost) with an uncolored phase.

b.i) The cubes that will have $3$ faces that are colored red are the cubes at the corner. There are a total of $8$ cubes at the corners.

b.ii) The cubes that will have $2$ faces that are colored red are the ones at the perimeter of the cube (except the corner cubes). There are a total of $96$ such cubes.

b.iii) The cubes that will have $1$ face that is colored red are the inner cubes that are at the surface of each phases of the cube. There are $384$ such cubes.

b.iv) The inside cubes are not colored. There are $512$ such cubes.

c.i) The cubes that will have $3$ faces that are colored red are the cubes at the corner. There are a total of $8$ cubes at the corners.

c.ii) The cubes that will have $2$ faces that are colored red are the ones at the perimeter of the cube (except the corner cubes). There are a total of $12(n-2)$ such cubes.

c.iii) The cubes that will have $1$ face that is colored red are the inner cubes that are at the surface of each phases of the cube. There are $6(n-2)^2$ such cubes.

c.iv) The inside cubes are not colored. There are $(n-2)^3$ such cubes.

d.i) There are a total of $8$ that will have $3$ phases that are colored red. If $n=3$ or $n=10,$ this, remains constant.

d.ii) There are a total of $12(n-2)$ cubes that have $2$ faces that are colored red.

$$
begin{align*}
text{If $n=3:$ }&
12(3-2)
\&=
12(1)
\&=
12
\\text{If $n=10:$ }&
12(10-2)
\&=
12(8)
\&=
96
.end{align*}
$$

d.iii) There are $6(n-2)^2$ cubes that will have $1$ face that is colored red.

$$
begin{align*}
text{If $n=3:$ }&
6(3-2)^2
\&=
6(1)^2
\&=
6(1)
\&=
6
\\text{If $n=10:$ }&
6(10-2)^2
\&=
6(8)^2
\&=
6(64)
\&=
384
.end{align*}
$$

d.iv) There are $(n-2)^3$ inside cubes which are not colored.

$$
begin{align*}
text{If $n=3:$ }&
(3-2)^3
\&=
(1)^3
\&=
1
\\text{If $n=10:$ }&
(10-2)^3
\&=
(8)^3
\&=
512
.end{align*}
$$

$$
begin{align*}
&text{a.i) } 8 && text{a.ii) } 12 && text{a.iii) } 6 && text{a.iv) } 1
\&text{b.i) } 8 && text{b.ii) } 96 && text{b.iii) } 384 && text{b.iv) } 512
\&text{c.i) } 8 && text{c.ii) } 12(n-2) && text{c.iii) } 6(n-2)^2 && text{c.iv) } (n-2)^3
\&text{d) see verification}
.end{align*}
$$

a) Let the two digit number be $45.$ Then

Step i) $4$

Step ii) $4+4^2$ equal to $20$

Step iii) $2025$

Hence, $45^2$ is equal to $2025.$

b) Let $a$ be the tens digit of a two-digit number ending in $5.$ Then

Step i) $a$

Step ii) $a+a^2$

Step iii) $100(a+a^2)+25$

The method above works all the time because the two-digit number ending in $5$, $underline{a}text{ }underline{5},$ is equal to $10a+5.$ When this number is squared, then

$$
begin{align*}
&
(10a+5)^2
\&=
(10a)^2+2(10a)(5)+5^2
\&=
100a^2+100a+25
\&=
100(a^2+a)+25
\&=
100(a+a^2)+25
.end{align*}
$$

The expression above is equivalent to Step iii.