Complete NCERT solutions for class 8 Maths Chapter 9 Algebraic Expressions

1.Identify the terms, their coefficients for each of the following expressions :
(i) $5xy{z}^2–3zy$

(ii) $1+x+x^2$

(iii) $4x^2{y}^2–4x^2{y}^2{z}^2+z^2$

(iv) $3–pq+qr?rp$

(v) $\frac{x}{2}+\frac{y}{2}?xy$

(vi)$0.3a–0.6ab+0.5b$

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Answer :

(i) We have the expression $5xy{z}^2–3zy$, the terms are $5xy{z}^2$ and $?3zy$.
Coefficient of $xy{z}^2$ in the term $5xy{z}^2$ is 5.

Coefficient of $zy$ in the term $–3yz$ is $–3$.

(ii)We have the expression $1+x+x^2$, the terms are 1, x and $x^2$.

Coefficient of the term 1 is 1.

Coefficient of x in the term x is 1.

Coefficient of $x^2$ in the term $x^2$ is 1.

(iii) We have the expression $4xy–4xyz+z$ , the terms are $4x^2{y}^2,–4x^2{y}^2{z}^2$ and $z^2$.

Coefficient of $x^2{y}^2$ in the term $4x^2{y}^2$ is 4.

Coefficient of $x^2{y}^2{z}^2$ in the term $–4x^2{y}^2{z}^2$ is – 4.

Coefficient of $z^2$ in the term $z^2$ is 1.

(iv) We have the expression $3–pq+qr–rp$, the terms are 3, – pq, qr and – rp.

Coefficient of the term 3 is 3.

Coefficient of pq in the term – pq is -1.

Coefficient of qr in the term qr is 1.

Coefficient of rp in the term – rp is -1.

(v) We have the expression $\frac{x}{2}+\frac{y}{2}?xy$, the terms are $\frac{x}{2},\frac{7}{2}$ and $–xy$.

Coefficient of x in the term $\frac{x}{2}$ is $\frac{1}{2}$

Coefficient of y in the term $\frac{y}{2}$ is $\frac{1}{2}$

Coefficient of xy in the term $–xy$ is -1.

(vi) In the expression $0.3a?0.6ab+0.5b$, the terms are $0.3a,–006ab$and $0.5b$.

Coefficient of a in the term 0.3a is 0.3.

Coefficient of ab in the term – 0.6ab is – 0.6.

Coefficient of b in the term 0.5b is 0.5.

2.Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories? $x+y,1000,x+x^2+x^3+x^4,7+y+5x,2y–3y^2,2y–3y^2+4y^3$,$5x–4y+3xy,4z–15z^2,ab+bc+cd+da,pqr,p^{2}q+p{q}^2,2p+2q.$

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Answer :

The calssification of the given expressions are:

Monomials : 1000, pqr

Binomials : $x+y,2y–3y^2,4z?15z^2,p^{2}q+p{q}^2,2p+2q.$

Trinomials : $7+y+5x,2y–3y^2+4y^3,5x–4y+3x.$

And the polynomials which don't have any category:

$x+x^2+x^3+x^4,ab+be+cd+da.$

3.Add the following :
(i) $ab–be,be–ca,ea–ab$

(ii)$a–b+ab,b?c+be,c–a+ac$

(iii) $2p^2{q}^2–3pq+4,5+7pq–3p^2{q}^2$

(iv)$l^2+m^2,m^2+n^2,n^2+l^2,2lm+2mn+2nl$

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Answer :

(i) We have the expressions as:$ab–bc,bc–ca,ca–ab$
Adding them :

$(ab–bc)+(bc–ca)+(ca–ab)$

$?ab–bc+bc–ca+ca–ab$

$?(ab–ab)+(bc–bc)+(ca–ca)$ [Placing the like terms together]

$?0+0+0$

$?0$

(ii) We have the expressions as:$a–b+ab,b–c+bc,c–a+ac$

Adding these we have:

$(a–b+ab)+(b–c+bc)+(c–a+ac)$

$?a–b+ab+b–c+bc+c–a+ac$

$?(a–a)+(b–b)+(c–c)+ab+bc+ac$[Placing all the like terms together]

$?0+0+0+ab+bc+ac$

$?ab+bc+ac$

(iii)We have the expressions as:$2p^2{q}^2–3pq+4,5+7pq–3p^2{q}^2$

Adding these we get: $2p^2{q}^2–3pq+4+5+7pq–3p^2{q}^2$

$?(2p^2{q}^2–3p^2{q}^2)+(–3pq+7pq)+(4+5)$[Placing like terms together]

$??p^2{q}^2+4pq+9$

(iv)We have the expressions as:$l^2+m^2,m^2+n^2,n^2+l^2,2lm+2mn+2nl$
Adding these we get: $(l^2+m^2)+(m^2+n^2)+(n^2+l^2)+(2lm+2mn+2nl)$

$?(l^2+l^2)+(m^2+m^2)+(n^2+n^2)+(2lm+2mn+2nl)$[Placing like terms together]

$?(2l^2)+(2m^2)+(2n^2)+(2lm+2mn+2nl$

$?2(l^2+m^2+n^2+lm+mn+nl)$

4.(a) Subtract $4a–7ab+3b+12$ from $12a–9ab+5b–3$
(b) Subtract $3xy+5yz–7zx$ from $5xy–2yz–2zx+10xyz$

(c) Subtract $4p^{2}q–3pq+5p{q}^2–8p+7q–10$ from $18–3p–11q+5pq–2p{q}^2+5p^{2}q$

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Answer :

(a)We have the expressions to subtract:$4a–7ab+3b+12$ from $12a–9ab+5b–3$
Subtracting these we get: $(12a–9ab+5b–3)?(4a–7ab+3b+12)$

$(12a–9ab+5b–3)+(?4a+7ab?3b?12)$[Taking (-) inside the second bracket]

$?(12a?4a)+(–9ab+7ab)+(+5b?3b)+(–3?12)$[Placing like terms together]

$?(8a)+(?2ab)+(2b)+(?15)$

$?(8a?2ab+2b?15)$

(b)We have the expressions to subtract:$3xy+5yz–7zx$ from $5xy–2yz–2zx+10xyz$
Subtracting these we get: $(5xy–2yz–2zx+10xyz)?(3xy+5yz–7zx)$

$(5xy–2yz–2zx+10xyz)+(?3xy?5yz+7zx)$[Taking (-) inside the second bracket]

$?(5xy?3xy)+(–2yz?5yz)+(–2zx+7zx)+(+10xyz)$[Placing like terms together]

$?(2xy)+(?7yz)+(5zx)+(10xyz)$

$?(2xy?7yz+5zx+10xyz)$

(c)We have the expressions to subtract:$4p^{2}q–3pq+5p{q}^2–8p+7q–10$ from $18–3p–11q+5pq–2p{q}^2+5p^{2}q$
Subtracting these we get: $(18–3p–11q+5pq–2p{q}^2+5p^{2}q)?(4p^{2}q–3pq+5p{q}^2–8p+7q–10)$

$?(18–3p–11q+5pq–2p{q}^2+5p^{2}q)+(?4p^{2}q+3pq?5p{q}^2+8p?7q+10)$[Taking (-) inside the second bracket]

$?(+18+10)+(?3p+8p)+(?11q?7q)+(5pq+3pq)+(?2p{q}^2?5p{q}^2)+(+5p^{2}q?4p^{2}q)$[Placing like terms together]

$?(+28)+(+5p)+(?18q)+(8pq)+(?7p{q}^2)+(+p^{2}q)$

$?(+28+5p?18q+8pq?7p{q}^2+p^{2}q)$

4.Obtain the volume of rectangular boxes with the following length, breadth and height respectively.
(i) $5a,3a^2,7a^4$

(ii) $2p,4q,8r$

(iii) $xy,2x^{2}y,2x{y}^2$

(iv) $a,2b,3c$

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Answer :

(i)Given that, length = 5a, breadth = $3a^2$, height = $7a^4$

?Volume of the box =$l\times b\times h=5a\times 3a^2\times 7a^4=105a^7$cu. units

(ii)Given that, length = 2p, breadth = 4q, height = 8r

?Volume of the box =$l\times b\times h=2p\times 4q\times 8r=64pqr$ cu. units

(iii)Given that, length = xy, breadth =$2x^{2}y$, height = 2xy^2\)

Volume of the box =$l\times b\times h=xy\times 2x^{2}y\times 2x{y}^2=(1\times 2\times 2)\times xy\times x^{2}y\times x{y}^2=4x^4{y}^4$cu. units

(iv)Given that, length = a, breadth = 2b, height = 3c

?Volume of the box = $l\times b\times h=a\times 2b\times 3c=(1\times 2\times 3)abc=6abc$ cu. units

1.Find the product of the following pairs of monomials :
(i) 4, 7p

(ii) -4p, 7p

(iii) – 4p, 7pq

(iv) 4p3 , – 3p

(v) 4p, 0

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Answer :

(i)Multiplying the given expressions: $4x\times 7p=(4\times 7)\times p=28p$

(ii) Multiplying the given expressions:$–4p\times 7p=(?4\times 7)\times (p\times p)$

$=?28p^{(1+1)}=–28p^2$

(iii)Multiplying the given expressions: $?4p\times 7pq=(?4\times 7)\times (p\times p\times q)$

$=?28p^{(1+1)}q=–28p^{2}q$

(iv)Multiplying the given expressions: $4p3\times –3p=(4\times –3)\times (p^3\times p)$

$=–12p^{(3+1)}==?12p^4$

(v) Multiplying the given expressions:$4p\times 0=(4\times 0)\times p=0\times p=0$

2.Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.
$(p,q);(10m,5n);(20x^2,5y^2);(4x,3x^2);(3mn,4np)$

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Answer :

(i) Given that: Length = p units and breadth = q units

Area of the rectangle = $length\times breadth=p\times q=pq$ sq units

(ii)Given that: Length = 10 m units, breadth = 5n units

?Area of the rectangle =$length\times breadth=10m\times 5n=(10\times 5)\times m\times n=50mn$sq units

(iii) Given that: Length = 20x2 units, breadth = 5y2 units

?Area of the rectangle = length × breadth = 20x2 × 5y2 = (20 × 5) × x2 × y2 = 100x2y2 sq units

(iv)Given that: Length = 4x units, breadth = 3x2 units

?Area of the rectangle = $length\times breadth=4x\times 3x^2=(4\times 3)\times x\times x^2=12x^3$ sq units

(v) Given that: Length = 3mn units, breadth = 4np units

?Area of the rectangle = $length\times breadth=3mn\times 4np=(3\times 4)\times mn\times np=12m{n}^{2}p$ sq units

3.Complete the table of products :

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Answer :

Completed table is as under :

5. Obtain the product of

(i) xy, yz, zx

(ii) a, -a2, a3

(iii) 2, 4y, 8y2, 16y3

(iv) a, 2b, 3c, 6abc

(v) m, -mn, mnp

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Answer :

(i) Product=$xy\times yz\times zx=x^2{y}^2{z}^2$

(ii) Product=$a\times (?a^2)\times a^3=(?a{)}^6$

(iii)Product=$2\times 4y\times 8y^2\times 16y^3=(2\times 4\times 8\times 16)\times y\times y^2\times y^3=1024y^6$

(iv)Product=$a\times 2b\times 3c\times 6abc=(1\times 2\times 3\times 6)\times a\times b\times c\times abc=36a^2{b}^2{c}^2$

(v)Product=$m\times (?mn)\times mnp=[1\times (?1)\times 1]m\times mn\times mnp=?m^3{n}^{2}p$

1.Carry out the multiplication of the expressions in each of the following pairs:
(i) $4p,q+r$

(ii) $ab,a–b$

(iii) $a+b,7a^2{b}^2$

(iv) $a^2–9,4a$

(v) $pq+qr+rp,0$

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Answer :

Multiplication of the expressions are as follows:
(i) $4p\times (q+r)=(4p\times q)+(4p\times r)=4pq+4pr$

(ii) $ab,a–b=ab\times (a–b)=(ab\times a)–(ab\times b)=a^{2}b–a{b}^2$

(iii) $(a+b)\times 7a^2{b}^2=(a\times 7a^2{b}^2)+(b\times 7a^2{b}^2)=7a^3{b}^2+7a^2{b}^3$

(iv) $(a2–9)\times 4a=(a^2\times 4a)–(9\times 4a)=4a^3–36a$

(v) $(pq+qr+rp)\times 0=0$

2. Complete the table.

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Answer :

Products of the given expressions in the table are done as follows:
.(i) $a\times (b+c+d)=(a\times b)+(a\times c)+(a\times d)=ab+ac+ad$

(ii) $(x+y–5)(5xy)=(x\times 5xy)+(y\times 5xy)–(5\times 5xy)=5x^{2}y+5x{y}^2–25xy$

(iii) $p\times (6p^2–7p+5)=(p\times 6p^2)–(p\times 7p)+(p\times 5)=6p^3–7p^2+5p$

(iv) $4p^2{q}^2\times (p^2–q^2)=4p^2{q}^2\times p^2–4p^2{q}^2\times q^2=4p^4{q}^2–4p^2{q}^4$

(v) $(a+b+c)\times (abc)=(a\times abc)+(b\times abc)+(c\times abc)=a^{2}bc+a{b}^{2}c+ab{c}^2$

Therefore, the completed table is as follows:

3.Find the products
(i)$(a^2)\times (2a^{22})\times (4a^{26})$

(ii)$(\frac{2}{3}xy)\times (?\frac{9}{10}{x}^2{y}^2)$

(iii)$(?\frac{10}{3}p{q}^3)\times (\frac{6}{5}{p}^{3}q)$

(iv)$x\times x^2\times x^3\times x^4$

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Answer :

(i)$(a^2)\times (2a^{22})\times (4a^{26})$

$=(1\times 2\times 4)\times a^{2+22+26}=8a^{50}$

(ii)$(\frac{2}{3}xy)\times (?\frac{9}{10}{x}^2{y}^2)$

$=(\frac{2}{3}\times (?\frac{9}{10}))\times x^{(1+2})y^{(1+2)}$

(iii)$(?\frac{10}{3}p{q}^3)\times (\frac{6}{5}{p}^{3}q)$

$=(?\frac{10}{3}\times \frac{6}{5})\times p^{(1+3)}{q}^{(1+3)}$

$=?4p^4{q}^4$

(iv)$x\times x^2\times x^3\times x^4$

\(= x^{(1+2+3+4)}=x^{10\)

4.(a) Simplify: $3x(4x–5)+3$ and find its values for (i) x = 3 (ii) x = $\frac{1}{2}$.
(b) Simplify: $a(a^2+a+1)+5$ and find its value for (i) a = 0 (ii) a = 1 (iii) a = -1

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Answer :

(a) Given the expression: $3x(4x–5)+3=4x\times 3x–5\times 3x+3=12x^2–15x+3$
(i) So for x = 3, we have

$12\times (3)2–15\times 3+3=12\times 9–45+3=108–42=66$

(ii)For x=$\frac{1}{2}$, we have:

$?12(\frac{1}{2}{)}^2?15(\frac{1}{2})+3$

$?12\times (\frac{1}{4})?\frac{15}{2}+3$

$?3?\frac{15}{2}+3$

$?\frac{6?15+6}{2}=\frac{12?15}{2}=?\frac{3}{2}$

(b) We have $a(a^2+a+1)+5$

$=(a^2\times a)+(a\times a)+(1\times a)+5$

$=a^3+a^2+a+5$

(i) For a = 0, we have:

$=(0{)}^3+(0{)}^2+(0)+5=5$

(ii) For a = 1, we have:

$=(1{)}^3+(1{)}^2+(1)+5=1+1+1+5=8$

(iii) For a = -1, we have:

$=(?1{)}^3+(?1{)}^2+(?1)+5=?1+1–1+5=4$

5.(a) $Add:p(p–q),q(q–r)andr(r–p)$
(b) $Add:2x(z–x–y)and2y(z–y–x)$

(c) $Subtract:3l(l–4m+5n)from4l(10n–3m+2l)$

(d) $Subtract:3a(a+b+c)–2b(a–b+c)from4c(?a+b+c)$

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Answer :

(a) $p(p–q)+q(q–r)+r(r–p)$

=$(p\times p)–(p\times q)+(q\times q)–(q\times r)+(r\times r)–(r\times p)$[Opening the brackets]

= $p^2–pq+q^2–qr+r^2–rp$

=$p^2+q^2+r^2–pq–qr–rp$

(b) $2x(z–x–y)+2y(z–y–x)$

= $(2x\times z)–(2x\times x)–(2x\times y)+(2y\times z)–(2y\times y)–(2y\times x)$[Opening the brackets]

= $2xz–2x^2–2xy+2yz–2y^2–2xy$

$=?2x^2–2y^2+2xz+2yz–4xy$

$=?2x^2–2y^2–4xy+2yz+2xz$

(c) $4l(10n–3m+2l)–3l(l–4m+5n)$

$=(4l\times 10n)–(4l\times 3m)+(4l\times 2l)–(3l\times l)–(3l\times ?4m)–(3l\times \times 5n)$[Opening the brackets]

$=40ln–12lm+8l^2–3l^2+12lm–15ln$

$=(40ln–15ln)+(?12lm+12lm)+(8l^2–3l^2)$

$=25ln+0+5l^2$

$=25ln+5l^2$

$=5l^2+25ln$

(d)$[4c(?a+b+c)]–[3a(a+b+c)–2b(a–b+c)]$

$=(?4ac+4bc+4c^2)–(3a^2+3ab+3ac–2ab+2b^2–2bc)$[Opening the brackets]

$=?4ac+4bc+4c^2–3a^2–3ab–3ac+2ab–2b^2+2bc$

$=?3a^2–2b^2+4c^2–ab+6bc–7ac$

1.Multiply the binomials:
(i) $(2x+5)$and $(4x–3)$

(ii) $(y–8)$ and $(3y–4)$

(iii) $(2.5l–0.5m)$ and $(2.5l+0.5m)$

(iv) $(a+3b)$ and $(x+5)$

(v) $(2pq+3q^2)$ and $(3pq–2q^2)$

(vi) $(\frac{3}{4}{a}^2+3b^2)$ and $4(a^2–\frac{2}{3}{b}^2)$

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Answer :

(i)Multiplying the given expressions as:$(2x+5)\times (4x–3)$

$=2x\times (4x–3)+5\times (4x–3)$

$=(2x\times 4x)–(3\times 2x)+(5\times 4x)–(5\times 3)$

$=8x^2–6x+20x–15$

$=8x^2+14x–15$

(ii)Multiplying the given expressions as: $(y–8)\times (3y–4)$

$=y\times (3y–4)–8\times (3y–4)$

$=(y\times 3y)–(y\times 4)–(8\times 3y)+(?8\times ?4)$

$=3y^2–4y–24y+32$

$=3y^2–28y+32$

(iii)Multiplying the given expressions as: $(2.5l–0.5m)\times (2.5l+0.5m)$

$=(2.5l\times 2.5l)+(2.5l\times 0.5m)–(0.5m\times 2.5l)–(0.5m\times 0.5m)$

$=6.25l^2+1.25ml–1.25ml–0.25m^2$

$=6.25l^2+0–0.25m^2$

$=6.25l^2–0.25m^2$

(iv)Multiplying the given expressions as: $(a+3b)\times (x+5)$

$=a\times (x+5)+36\times (x+5)$

$=(a\times x)+(a\times 5)+(36\times x)+(36\times 5)$

$=ax+5a+3bx+15b$

(v) Multiplying the given expressions as:$(2pq+3q^2)\times (3pq–2q^2)$

$=2pq\times (3pq–2q^2)+3q^2(3pq–2q^2)$

$=(2pq\times 3pq)–(2pq\times 2q^2)+(3q^2\times 3pq)–(3q^2\times 2q^2)$

$=6p^2{q}^2–4p{q}^3+9p{q}^3–6q^4$

$=6p^2{q}^2+5p{q}^3–6q^4$

(vi)Multiplying the given expressions as:$(\frac{3}{4}{a}^2+3b^2)\times 4(a^2–\frac{2}{3}{b}^2)$

$=(\frac{3}{4}{a}^2+3b^2)\times (4a^2?\frac{8}{3}{b}^2)$

$=\frac{3}{4}{a}^2\times (4a^2?\frac{8}{3}{b}^2)+3b^2\times (4a^2?\frac{8}{3}{b}^2)$

$=(\frac{3}{4}{a}^2\times 4b^2)?(\frac{3}{4}{a}^2\times \frac{8}{3}{b}^2)+(3b^2\times 4a^2)?(3b^2\times \frac{8}{3}{b}^2)$

$=3a^4?2a^2{b}^2+12a^2{b}^2?8b^4$

$=3a^4+10a^2{b}^2?8b^4$

2.Find the product:
(i) $(5–2x)(3+x)$

(ii)$(x+7y)(7x–y)$

(iii) $(a^2+b)(a+b^2)$

(iv)$(p^2–q^2)(2p+q)$

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Answer :

(i)Product of the expression:$(5–2x)(3+x)$

$=5(3+x)–2x(3+x)$

$=(5\times 3)+(5\times x)–(2x\times 3)–(2x\times x)$

$=15+5x–6x–2x^2$

(ii)Product of the expression: $(x+7y)(7x–y)$

$=x(7x–y)+7y(7x–y)$

$=(x\times 7x)–(x\times y)+(7y\times 7x)–(7y\times y)$

$=7x^2–xy+49xy–7y^2$

$=7x^2+48xy–7y^2$

(iii)Product of the expression: $(a^2+b)(a+b^2)$

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

$=(a^2\times a)+(a^2\times b^2)+(b\times a)+(b\times b^2)$

$=a^3+a^2{b}^2+ab+b^3$

(iv)Product of the expression:$(p^2–q^2)(2p+q)$

$=p^2(2p+q)–q^2(2p+q)$

$=(p^2\times 2p)+(p^2\times q)–(q^2\times 2p)–(q^2\times q)$

$=2p^3+p^{2}q–2p{q}^2–q^3$

3.Simplify: (i) $(x^2–5)(x+5)+25$

(ii) $(a^2+5)(b^3+3)+5$

(iii) $(t+s^2)(t^2–s)$

(iv)$(a+b)(c–d)+(a–b)(c+d)+2(ac+bd)$

(v)$(x+y)(2x+y)+(x+2y)(x–y)$

(vi) $(x+y)(x^2–xy+y^2)$

(vii)$(1.5x–4y)(1.5x+4y+3)–4.5x+12y$

(viii)$(a+b+c)(a+b–c)$

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Answer :

(i)Simplifying the expression, we have: $(x^2–5)(x+5)+25$

$=x^2\times (x+5)+5\times (x+5)+25$

$=x^3+5x^2–5x–25+25$

$=x^3+5x^2–5x+0$

$=x^3+5x^2–5x$

(ii)Simplifying the expression, we have: $(a^2+5)(b^3+3)+5$

$=a^2\times (b^3+3)+5\times (b^3+3)+5$

$=a^2{b}^3+3a^2+5b^3+15+5$

$=a^2{b}^3+3a^2+5b^3+20$

(iii)Simplifying the expression, we have:$(t+s^2)(t^2–s)$

$=t\times (t^2–s)+s^2\times (t^2–s)$

$=t^3–st+s^2{t}^2–s^3$

$=t^3+s^2{t}^2–st–s^3$

(iv) Simplifying the expression, we have:$(a+b)(c–d)+(a–b)(c+d)+2(ac+bd)$

$=a\times (c–d)+b\times (c–d)+a\times (c+d)–b\times (c+d)+2ac+2bd$

$=ac–ad+bc–bd+ac+ad–bc–bd+2ac+2bd$

$=ac+ac+2ac+bc–bc–ad+ad–bd–bd+2bd$

$=4ac+0+0+0$

$=4ac$

(v) Simplifying the expression, we have:$(x+y)(2x+y)+(x+2y)(x–y)$

$=x\times (2x+y)+y\times (2x+y)+x\times (x–y)+2y\times (x–y)$

$=2x^2+xy+2xy+y^2+x^2–xy+2xy–2y^2$

$=2x^2+x^2+xy+2xy–xy+2xy+y^2–2y^2$

$=3x^2+4xy–y^2$

(vi)Simplifying the expression, we have:$(x+y)(x^2–xy+y^2)$

$=x\times (x^2–xy+y^2)+y\times (x^2–xy+y^2)$

$=x^3–x^{2}y+x^{2}y+x{y}^2–x{y}^2+y^3$

$=x^3–0+0+y^3$

$=x^3+y^3$

(vii)Simplifying the expression, we have:$(1.5x–4y)(1.5x+4y+3)–4.5x.+12y$

$=1.5x\times (1.5x+4y+3)–4y\times (1.5x+4y+3)–4.5x+12y$

$=2.25x^2+6xy+4.5x–6xy–16y^2–12y–4.5x+12y$

$=2.25x^2+6xy–6xy+4.5x–4.5x+12y–12y–16y^2$

$=2.25x^2+0+0+0–16y62$

$=2.25x^2–16y^2$

(viii)Simplifying the expression, we have: $(a+b+c)(a+b–c)$

$=a(a+b–c)+b(a+b–c)+c(a+b–c)$

$=a^2+ab–ac+ab+b^2–bc+ac+bc–c^2$

$=a^2+ab+ab–bc+bc–ac+ac+b^2–c^2$

$=a^2+2ab+b^2–c^2+0+0$

$=a^2+2ab+b^2–c^2$