# NCERT Solutions for Class 8 Math Chapter 9 – Algebraic Expressions and Identities

#### Question 1:

Identify the terms, their coefficients for each of the following expressions.

(i) 5xyz2 − 3zy

(ii) 1 + x + x2

(iii) 4x2y2 − 4x2y2z2 + z2

(iv) 3 − pq + qrrp

(v)

(vi) 0.3a − 0.6ab + 0.5b

The terms and the respective coefficients of the given expressions are as follows.

 – Terms Coefficients (i) 5xyz2− 3zy 5− 3 (ii) 1xx2 111 (iii) 4x2y2− 4x2y2z2z2 4− 41 (iv) 3− pqqr− rp 3−11−1 (v) − xy − 1 (vi) 0.3a− 0.6ab0.5b 0.3− 0.60.5

#### Question 2:

Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

x + y, 1000, x + x2 + x3 + x4, 7 + y + 5x, 2y − 3y2, 2y − 3y2 + 4y3, 5x − 4y + 3xy, 4z − 15z2, ab + bc + cd + da, pqr, p2q + pq2, 2p + 2q

The given expressions are classified as

Monomials: 1000, pqr

Binomials: x + y, 2y − 3y2, 4z − 15z2, p2q + pq2, 2p + 2q

Trinomials: 7 + y + 5x, 2y − 3y2 + 4y3, 5x − 4y + 3xy

Polynomials that do not fit in any of these categories are

x + x2 + x3 + x4, ab + bc + cd + da

#### Question 3:

(i) abbc, bcca, caab

(ii) ab + ab, bc + bc, ca + ac

(iii) 2p2q2 − 3pq + 4, 5 + 7pq − 3p2q2

(iv) l2 + m2, m2 + n2, n2 + l2, 2lm + 2mn + 2nl

The given expressions written in separate rows, with like terms one below the other and then the addition of these expressions are as follows.

(i)

Thus, the sum of the given expressions is 0.

(ii)

Thus, the sum of the given expressions is ab + bc + ac.

(iii)

Thus, the sum of the given expressions is −p2q2 + 4pq + 9.

(iv)

Thus, the sum of the given expressions is 2(l2 + m2 + n2 + lm + mn + nl).

#### Question 4:

(a) Subtract 4a − 7ab + 3b + 12 from 12a − 9ab + 5b − 3

(b) Subtract 3xy + 5yz − 7zx from 5xy − 2yz − 2zx + 10xyz

(c) Subtract 4p2q − 3pq + 5pq2 − 8p + 7q − 10 from 18 − 3p − 11q + 5pq − 2pq2 + 5p2q

The given expressions in separate rows, with like terms one below the other and then the subtraction of these expressions is as follows.

(a)

(b)

(c)

#### Question 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

The product will be as follows.

(i) 4 × 7p = 4 × 7 × p = 28p

(ii) − 4p × 7p = − 4 × p × 7 × p = (− 4 × 7) × (p × p) = − 28 p2

(iii) − 4p × 7pq = − 4 × p × 7 × p × q = (− 4 × 7) × (p × p × q) = − 28p2q

(iv) 4p3 × − 3p = 4 × (− 3) × p × p × p × p = − 12 p4

(v) 4p × 0 = 4 × p × 0 = 0

#### Question 2:

Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.

(p, q); (10m, 5n); (20x2, 5y2); (4x, 3x2); (3mn, 4np)

We know that,

Area of rectangle = Length × Breadth

Area of 1st rectangle = p × q = pq

Area of 2nd rectangle = 10m × 5n = 10 × 5 × m × n = 50 mn

Area of 3rd rectangle = 20x2 × 5y2 = 20 × 5 × x2 × y2 = 100 x2y2

Area of 4th rectangle = 4x × 3x2 = 4 × 3 × x × x2 = 12x3

Area of 5th rectangle = 3mn × 4np = 3 × 4 × m × n × n × p = 12mn2p

#### Question 3:

Complete the table of products.

 2x − 5y 3x2 − 4xy 7x2y − 9x2y2 2x 4x2 … … … … … − 5y … … − 15x2y … … … 3x2 … … … … … … − 4xy … … … … … … 7x2y … … … … … … − 9x2y2 … … … … … …

The table can be completed as follows.

 2x − 5y 3x2 − 4xy 7x2y − 9x2y2 2x 4x2 − 10xy 6x3 − 8x2y 14x3y − 18x3y2 − 5y − 10xy 25 y2 − 15x2y 20xy2 − 35x2y2 45x2y3 3x2 6x3 − 15x2y 9x4 − 12x3y 21x4y − 27x4y2 − 4xy − 8x2y 20xy2 − 12x3y 16x2y2 − 28x3y2 36x3y3 7x2y 14x3y − 35x2y2 21x4y − 28x3y2 49x4y2 − 63x4y3 − 9x2y2 − 18x3y2 45 x2y3 − 27x4y2 36x3y3 − 63x4y3 81x4y4

#### Question 4:

Obtain the volume of rectangular boxes with the following length, breadth and height respectively.

(i) 5a, 3a2, 7a4 (ii) 2p, 4q, 8r (iii) xy, 2x2y, 2xy2

(iv) a, 2b, 3c

We know that,

Volume = Length × Breadth × Height

(i) Volume = 5a × 3a2 × 7a4 = 5 × 3 × 7 × a × a2 × a4 = 105 a7

(ii) Volume = 2p × 4q × 8r = 2 × 4 × 8 × p × q × r = 64pqr

(iii) Volume = xy × 2x2y × 2xy2 = 2 × 2 × xy ×x2y × xy2 = 4x4y4

(iv) Volume = a × 2b × 3c = 2 × 3 × a × b × c = 6abc

#### Question 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

(i) xy × yz × zx = x2y2z2

(ii) a × (− a2) × a3 = − a6

(iii) 2 × 4y × 8y2 × 16y3 = 2 × 4 × 8 × 16 × y × y2 × y3 = 1024 y6

(iv) a × 2b × 3c × 6abc = 2 × 3 × 6 × a × b × c × abc = 36a2b2c2

(v) m × (− mn) × mnp = − m3n2p

#### Question 1:

Carry out the multiplication of the expressions in each of the following pairs.

(i) 4p, q + r (ii) ab, ab (iii) a + b, 7a2b2

(iv) a2 − 9, 4a (v) pq + qr + rp, 0

(i) (4p) × (q + r) = (4p × q) + (4p × r) = 4pq + 4pr

(ii) (ab) × (ab) = (ab × a) + [ab × (− b)] = a2bab2

(iii) (a + b) × (7a2 b2) = (a × 7a2b2) + (b × 7a2b2) = 7a3b2 + 7a2b3

(iv) (a2 − 9) × (4a) = (a2 × 4a) + (− 9) × (4a) = 4a3 − 36a

(v) (pq + qr + rp) × 0 = (pq × 0) + (qr × 0) + (rp × 0) = 0

#### Question 2:

Complete the table

 — First expression Second Expression Product (i) a b + c + d – (ii) x + y − 5 5 xy – (iii) p 6p2 − 7p + 5 – (iv) 4p2q2 p2 − q2 – (v) a + b + c abc –

The table can be completed as follows.

 – First expression Second Expression Product (i) a b + c + d ab + ac + ad (ii) x + y − 5 5 xy 5x2y + 5xy2 − 25xy (iii) p 6p2 − 7p + 5 6p3 − 7p2 + 5p (iv) 4p2q2 p2 − q2 4p4q2 − 4p2q4 (v) a + b + c abc a2bc + ab2c + abc2

#### Question 3:

Find the product.

(i) (a2) × (2a22) × (4a26)

(ii)

(iii)

(iv) x × x2 × x3 × x4

(i) (a2) × (2a22) × (4a26) = 2 × 4 ×a2 × a22 × a26 = 8a50

(ii)

(iii)

(iv) x × x2 × x3 × x4 = x10

#### Question 4:

(a) Simplify 3x (4x −5) + 3 and find its values for (i) x = 3, (ii) .

(b) a (a2 + a + 1) + 5 and find its values for (i) a = 0, (ii) a = 1, (iii) a = − 1.

(a) 3x (4x − 5) + 3 = 12x2 − 15x + 3

(i) For x = 3, 12x2 − 15x + 3 = 12 (3)2 − 15(3) + 3

= 108 − 45 + 3

= 66

(ii) For

(b)a (a2 + a + 1) + 5 = a3 + a2 + a + 5

(i) For a = 0, a3 + a2 + a + 5 = 0 + 0 + 0 + 5 = 5

(ii) For a = 1, a3 + a2 + a + 5 = (1)3 + (1)2 + 1 + 5

= 1 + 1 + 1 + 5 = 8

(iii) For a = −1, a3 + a2 + a + 5 = (−1)3 + (−1)2 + (−1) + 5

= − 1 + 1 − 1 + 5 = 4

#### Question 5:

(a) Add: p (pq), q (q ­­­r) and r (r ­− p)

(b) Add: 2x (zxy) and 2y (zyx)

(c) Subtract: 3l (l − 4m + 5n) from 4l (10n − 3m + 2l)

(d) Subtract: 3a (a + b + c) − 2b (ab + c) from 4c (− a + b + c)

(a) First expression = p (pq) = p2pq

Second expression = q (q ­­­r) = q2qr

Third expression = r (r ­− p) = r2pr

Adding the three expressions, we obtain

Therefore, the sum of the given expressions is p2 + q2 + r2pqqrrp.

(b) First expression = 2x (zxy) = 2xz − 2x2 − 2xy

Second expression = 2y (zyx) = 2yz − 2y2 − 2yx

Adding the two expressions, we obtain

Therefore, the sum of the given expressions is − 2x2 − 2y2 − 4xy + 2yz + 2zx.

(c) 3l (l − 4m + 5n) = 3l2 − 12lm + 15ln

4l (10n − 3m + 2l) = 40ln − 12lm + 8l2

Subtracting these expressions, we obtain

Therefore, the result is 5l2 + 25ln.

(d) 3a (a + b + c) − 2b (ab + c) = 3a2 +3ab + 3ac − 2ba + 2b2 − 2bc

= 3a2 + 2b2 + ab + 3ac − 2bc

4c (− a + b + c) = − 4ac + 4bc + 4c2

Subtracting these expressions, we obtain

Therefore, the result is −3a2 −2b2 + 4c2ab + 6bc − 7ac.

#### Question 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 + 3q2) and (3pq − 2q2)

(vi)

(i) (2x + 5) × (4x − 3) = 2x × (4x − 3) + 5 × (4x − 3)

= 8x2 − 6x + 20x − 15

= 8x2 + 14x −15 (By adding like terms)

(ii) (y − 8) × (3y − 4) = y × (3y − 4) − 8 × (3y − 4)

= 3y2 − 4y − 24y + 32

= 3y2 − 28y + 32 (By adding like terms)

(iii) (2.5l − 0.5m) × (2.5l + 0.5m) = 2.5l × (2.5l + 0.5m) − 0.5m (2.5l + 0.5m)

= 6.25l2 + 1.25lm − 1.25lm − 0.25m2

= 6.25l2 − 0.25m2

(iv) (a + 3b) × (x + 5) = a × (x + 5) + 3b × (x + 5)

= ax + 5a + 3bx + 15b

(v) (2pq + 3q2) × (3pq − 2q2) = 2pq × (3pq − 2q2) + 3q2 × (3pq − 2q2)

= 6p2q2 − 4pq3 + 9pq3 − 6q4

= 6p2q2 + 5pq3 − 6q4

(vi)

#### Question 2:

Find the product.

(i) (5 − 2x) (3 + x) (ii) (x + 7y) (7xy)

(iii) (a2 + b) (a + b2) (iv) (p2q2) (2p + q)

(i) (5 − 2x) (3 + x) = 5 (3 + x) − 2x (3 + x)

= 15 + 5x − 6x − 2x2

= 15 − x − 2x2

(ii) (x + 7y) (7xy) = x (7xy) + 7y (7xy)

= 7x2xy + 49xy − 7y2

= 7x2 + 48xy − 7y2

(iii) (a2 + b) (a + b2) = a2 (a + b2) + b (a + b2)

= a3 + a2b2 + ab + b3

(iv) (p2q2) (2p + q) = p2 (2p + q) − q2 (2p + q)

= 2p3 + p2q − 2pq2q3

#### Question 3:

Simplify.

(i) (x2 − 5) (x + 5) + 25

(ii) (a2 + 5) (b3 + 3) + 5

(iii) (t + s2) (t2 − s)

(iv) (a + b) (cd) + (ab) (c + d) + 2 (ac + bd)

(v) (x + y) (2x + y) + (x + 2y) (xy)

(vi) (x + y) (x2xy + y2)

(vii) (1.5x − 4y) (1.5x + 4y + 3) − 4.5x + 12y

(viii) (a + b + c) (a + bc)

(i) (x2 − 5) (x + 5) + 25

= x2 (x + 5) − 5 (x + 5) + 25

= x3 + 5x2 − 5x − 25 + 25

= x3 + 5x2 − 5x

(ii) (a2 + 5) (b3 + 3) + 5

= a2 (b3 + 3) + 5 (b3 + 3) + 5

= a2b3 + 3a2 + 5b3 + 15 + 5

= a2b3 + 3a2 + 5b3 + 20

(iii) (t + s2) (t2 − s)

= t (t2 s) + s2 (t2 − s)

= t3st + s2t2 s3

(iv) (a + b) (cd) + (ab) (c + d) + 2 (ac + bd)

= a (cd) + b (cd) + a (c + d) − b (c + d) + 2 (ac + bd)

= acad + bcbd + ac + adbcbd + 2ac + 2bd

= (ac + ac + 2ac) + (adad) + (bcbc) + (2bdbdbd)

= 4ac

(v) (x + y) (2x + y) + (x + 2y) (xy)

= x (2x + y) + y (2x + y) + x (xy) + 2y (xy)

= 2x2 + xy + 2xy + y2 + x2xy + 2xy − 2y2

= (2x2 + x2) + (y2 − 2y2) + (xy + 2xyxy + 2xy)

= 3x2y2 + 4xy

(vi) (x + y) (x2xy + y2)

= x (x2xy + y2) + y (x2xy + y2)

= x3x2y + xy2 + x2yxy2 + y3

= x3 + y3 + (xy2xy2) + (x2yx2y)

= x3 + y3

(vii) (1.5x − 4y) (1.5x + 4y + 3) − 4.5x + 12y

= 1.5x (1.5x + 4y + 3) − 4y (1.5x + 4y + 3) − 4.5x + 12y

= 2.25 x2 + 6xy + 4.5x − 6xy − 16y2 − 12y − 4.5x + 12y

= 2.25 x2 + (6xy − 6xy) + (4.5x − 4.5x) − 16y2 + (12y − 12y)

= 2.25x2 − 16y2

(viii) (a + b + c) (a + bc)

= a (a + bc) + b (a + bc) + c (a + bc)

= a2 + abac + ab + b2bc + ca + bcc2

= a2 + b2c2 + (ab + ab) + (bcbc) + (caca)

= a2 + b2c2 + 2ab

#### Question 1:

Use a suitable identity to get each of the following products.

(i) (x + 3) (x + 3) (ii) (2y + 5) (2y + 5)

(iii) (2a ­− 7) (2a − 7) (iv)

(v) (1.1m − 0.4) (1.1 m + 0.4) (vi) (a2 + b2) (− a2 + b2)

(vii) (6x − 7) (6x + 7) (viii) (− a + c) (− a + c)

(ix) (x) (7a − 9b) (7a − 9b)

The products will be as follows.

(i) (x + 3) (x + 3) = (x + 3)2

= (x)2 + 2(x) (3) + (3)2 [(a + b)2 = a2 + 2ab + b2]

= x2 + 6x + 9

(ii) (2y + 5) (2y + 5) = (2y + 5)2

= (2y)2 + 2(2y) (5) + (5)2 [(a + b)2 = a2 + 2ab + b2]

= 4y2 + 20y + 25

(iii) (2a ­− 7) (2a − 7) = (2a − 7)2

= (2a)2 − 2(2a) (7) + (7)2 [(ab)2 = a2 − 2ab + b2]

= 4a2 − 28a + 49

(iv)

[(ab)2 = a2 − 2ab + b2]

(v) (1.1m − 0.4) (1.1 m + 0.4)

= (1.1m)2 − (0.4)2 [(a + b) (ab) = a2b2]

= 1.21m2 − 0.16

(vi) (a2 + b2) (− a2 + b2) = (b2 + a2) (b2a2)

= (b2)2 − (a2)2 [(a + b) (ab) = a2b2]

= b4a4

(vii) (6x − 7) (6x + 7) = (6x)2 − (7)2 [(a + b) (ab) = a2b2]

= 36x2 − 49

(viii) (− a + c) (− a + c) = (− a + c)2

= (− a)2 + 2(− a) (c) + (c)2 [(a + b)2 = a2 + 2ab + b2]

= a2 − 2ac + c2

(ix)

[(a + b)2 = a2 + 2ab + b2]

(x) (7a − 9b) (7a − 9b) = (7a − 9b)2

= (7a)2 − 2(7a)(9b) + (9b)2 [(a b)2 = a2 − 2ab + b2]

= 49a2 − 126ab + 81b2

#### Question 2:

Use the identity (x + a) (x + b) = x2 + (a + b)x + ab to find the following products.

(i) (x + 3) (x + 7) (ii) (4x +5) (4x + 1)

(iii) (4x − 5) (4x − 1) (iv) (4x + 5) (4x − 1)

(v) (2x +5y) (2x + 3y) (vi) (2a2 +9) (2a2 + 5)

(vii) (xyz − 4) (xyz − 2)

The products will be as follows.

(i) (x + 3) (x + 7) = x2 + (3 + 7) x + (3) (7)

= x2 + 10x + 21

(ii) (4x + 5) (4x + 1) = (4x)2 + (5 + 1) (4x) + (5) (1)

= 16x2 + 24x + 5

(iii)

= 16x2 − 24x + 5

(iv)

= 16x2 + 16x − 5

(v) (2x +5y) (2x + 3y) = (2x)2 + (5y + 3y) (2x) + (5y) (3y)

= 4x2 + 16xy + 15y2

(vi) (2a2 +9) (2a2 + 5) = (2a2)2 + (9 + 5) (2a2) + (9) (5)

= 4a4 + 28a2 + 45

(vii) (xyz − 4) (xyz − 2)

=

= x2y2z2 − 6xyz + 8

#### Question 3:

Find the following squares by suing the identities.

(i) (b − 7)2 (ii) (xy + 3z)2 (iii) (6x2 − 5y)2

(iv) (v) (0.4p − 0.5q)2 (vi) (2xy + 5y)2

(i) (b − 7)2 = (b)2 − 2(b) (7) + (7)2 [(ab)2 = a2 − 2ab + b2]

= b2 − 14b + 49

(ii) (xy + 3z)2 = (xy)2 + 2(xy) (3z) + (3z)2 [(a + b)2 = a2 + 2ab + b2]

= x2y2 + 6xyz + 9z2

(iii) (6x2 − 5y)2 = (6x2)2 − 2(6x2) (5y) + (5y)2 [(ab)2 = a2 − 2ab + b2]

= 36x4 − 60x2y + 25y2

(iv) [(a + b)2 = a2 + 2ab + b2]

(v) (0.4p − 0.5q)2 = (0.4p)2 − 2 (0.4p) (0.5q) + (0.5q)2

[(ab)2 = a2 − 2ab + b2]

= 0.16p2 − 0.4pq + 0.25q2

(vi) (2xy + 5y)2 = (2xy)2 + 2(2xy) (5y) + (5y)2

[(a + b)2 = a2 + 2ab + b2]

= 4x2y2 + 20xy2 + 25y2

#### Question 4:

Simplify.

(i) (a2b2)2 (ii) (2x +5)2 − (2x − 5)2

(iii) (7m − 8n)2 + (7m + 8n)2 (iv) (4m + 5n)2 + (5m + 4n)2

(v) (2.5p − 1.5q)2 − (1.5p − 2.5q)2

(vi) (ab + bc)2 − 2ab2c (vii) (m2n2m)2 + 2m3n2

(i) (a2b2)2 = (a2)2 − 2(a2) (b2) + (b2)2 [(ab)2 = a2 − 2ab + b2 ]

= a4 − 2a2b2 + b4

(ii) (2x +5)2 − (2x − 5)2 = (2x)2 + 2(2x) (5) + (5)2 − [(2x)2 − 2(2x) (5) + (5)2]

[(ab)2 = a2 − 2ab + b2]

[(a + b)2 = a2 + 2ab + b2]

= 4x2 + 20x + 25 − [4x2 − 20x + 25]

= 4x2 + 20x + 25 − 4x2 + 20x − 25 = 40x

(iii) (7m − 8n)2 + (7m + 8n)2

= (7m)2 − 2(7m) (8n) + (8n)2 + (7m)2 + 2(7m) (8n) + (8n)2

[(ab)2 = a2 − 2ab + b2 and (a + b)2 = a2 + 2ab + b2]

= 49m2 − 112mn + 64n2 + 49m2 + 112mn + 64n2

= 98m2 + 128n2

(iv) (4m + 5n)2 + (5m + 4n)2

= (4m)2 + 2(4m) (5n) + (5n)2 + (5m)2 + 2(5m) (4n) + (4n)2

[ (a + b)2 = a2 + 2ab + b2]

= 16m2 + 40mn + 25n2 + 25m2 + 40mn + 16n2

= 41m2 + 80mn + 41n2

(v) (2.5p − 1.5q)2 − (1.5p − 2.5q)2

= (2.5p)2 − 2(2.5p) (1.5q) + (1.5q)2 − [(1.5p)2 − 2(1.5p)(2.5q) + (2.5q)2]

[(ab)2 = a2 − 2ab + b2 ]

= 6.25p2 − 7.5pq + 2.25q2 − [2.25p2 − 7.5pq + 6.25q2]

= 6.25p2 − 7.5pq + 2.25q2 − 2.25p2 + 7.5pq − 6.25q2]

= 4p2 − 4q2

(vi) (ab + bc)2 − 2ab2c

= (ab)2 + 2(ab)(bc) + (bc)2 − 2ab2c [(a + b)2 = a2 + 2ab + b2 ]

= a2b2 + 2ab2c + b2c2 − 2ab2c

= a2b2 + b2c2

(vii) (m2n2m)2 + 2m3n2

= (m2)2 − 2(m2) (n2m) + (n2m)2 + 2m3n2 [(ab)2 = a2 − 2ab + b2 ]

= m4 − 2m3n2 + n4m2 + 2m3n2

= m4 + n4m2

#### Question 5:

Show that

(i) (3x + 7)2 − 84x = (3x − 7)2 (ii) (9p − 5q)2 + 180pq = (9p + 5q)2

(iii)

(iv) (4pq + 3q)2 − (4pq − 3q)2 = 48pq2

(v) (ab) (a + b) + (bc) (b + c) + (ca) (c + a) = 0

(i) L.H.S = (3x + 7)2 − 84x

= (3x)2 + 2(3x)(7) + (7)2 − 84x

= 9x2 + 42x + 49 − 84x

= 9x2 − 42x + 49

R.H.S = (3x − 7)2 = (3x)2 − 2(3x)(7) +(7)2

= 9x2 − 42x + 49

L.H.S = R.H.S

(ii) L.H.S = (9p − 5q)2 + 180pq

= (9p)2 − 2(9p)(5q) + (5q)2 − 180pq

= 81p2 − 90pq + 25q2 + 180pq

= 81p2 + 90pq + 25q2

R.H.S = (9p + 5q)2

= (9p)2 + 2(9p)(5q) + (5q)2

= 81p2 + 90pq + 25q2

L.H.S = R.H.S

(iii) L.H.S =

(iv) L.H.S = (4pq + 3q)2 − (4pq − 3q)2

= (4pq)2 + 2(4pq)(3q) + (3q)2 − [(4pq)2 − 2(4pq) (3q) + (3q)2]

= 16p2q2 + 24pq2 + 9q2 − [16p2q2 − 24pq2 + 9q2]

= 16p2q2 + 24pq2 + 9q2 −16p2q2 + 24pq2 − 9q2

= 48pq2 = R.H.S

(v) L.H.S = (ab) (a + b) + (bc) (b + c) + (ca) (c + a)

= (a2b2) + (b2c2) + (c2a2) = 0 = R.H.S.

#### Question 6:

Using identities, evaluate.

(i) 712 (ii) 992 (iii) 1022 (iv) 9982

(v) (5.2)2 (vi) 297 × 303 (vii) 78 × 82

(viii) 8.92 (ix) 1.05 × 9.5

(i) 712 = (70 + 1)2

= (70)2 + 2(70) (1) + (1)2 [(a + b)2 = a2 + 2ab + b2 ]

= 4900 + 140 + 1 = 5041

(ii) 992 = (100 − 1)2

= (100)2 − 2(100) (1) + (1)2 [(ab)2 = a2 − 2ab + b2 ]

= 10000 − 200 + 1 = 9801

(iii) 1022 = (100 + 2)2

= (100)2 + 2(100)(2) + (2)2 [(a + b)2 = a2 + 2ab + b2 ]

= 10000 + 400 + 4 = 10404

(iv) 9982 = (1000 − 2)2

= (1000)2 − 2(1000)(2) + (2)2 [(ab)2 = a2 − 2ab + b2 ]

= 1000000 − 4000 + 4 = 996004

(v) (5.2)2 = (5.0 + 0.2)2

= (5.0)2 + 2(5.0) (0.2) + (0.2)2 [(a + b)2 = a2 + 2ab + b2 ]

= 25 + 2 + 0.04 = 27.04

(vi) 297 × 303 = (300 − 3) × (300 + 3)

= (300)2 − (3)2 [(a + b) (ab) = a2b2]

= 90000 − 9 = 89991

(vii) 78 × 82 = (80 − 2) (80 + 2)

= (80)2 − (2)2 [(a + b) (ab) = a2b2]

= 6400 − 4 = 6396

(viii) 8.92 = (9.0 − 0.1)2

= (9.0)2 − 2(9.0) (0.1) + (0.1)2 [(ab)2 = a2 − 2ab + b2 ]

= 81 − 1.8 + 0.01 = 79.21

(ix) 1.05 × 9.5 = 1.05 × 0.95 × 10

= (1 + 0.05) (1− 0.05) ×10

= [(1)2 − (0.05)2] × 10

= [1 − 0.0025] × 10 [(a + b) (ab) = a2b2]

= 0.9975 × 10 = 9.975

#### Question 7:

Using a2 b2 = (a + b) (ab), find

(i) 512 − 492 (ii) (1.02)2 − (0.98)2 (iii) 1532 − 1472

(iv) 12.12 − 7.92

(i) 512 − 492 = (51 + 49) (51 − 49)

= (100) (2) = 200

(ii) (1.02)2 − (0.98)2 = (1.02 + 0.98) (1.02 ­− 0.98)

= (2) (0.04) = 0.08

(iii) 1532 − 1472 = (153 + 147) (153 − 147)

= (300) (6) = 1800

(iv) 12.12 − 7.92 = (12.1 + 7.9) (12.1 − 7.9)

= (20.0) (4.2) = 84

#### Question 8:

Using (x + a) (x + b) = x2 + (a + b) x + ab, find

(i) 103 × 104 (ii) 5.1 × 5.2 (iii) 103 × 98 (iv) 9.7 × 9.8

(i) 103 × 104 = (100 + 3) (100 + 4)

= (100)2 + (3 + 4) (100) + (3) (4)

= 10000 + 700 + 12 = 10712

(ii) 5.1 × 5.2 = (5 + 0.1) (5 + 0.2)

= (5)2 + (0.1 + 0.2) (5) + (0.1) (0.2)

= 25 + 1.5 + 0.02 = 26.52

(iii) 103 × 98 = (100 + 3) (100 − 2)

= (100)2 + [3 + (− 2)] (100) + (3) (− 2)

= 10000 + 100 − 6

= 10094

(iv) 9.7 × 9.8 = (10 − 0.3) (10 − 0.2)

= (10)2 + [(− 0.3) + (− 0.2)] (10) + (− 0.3) (− 0.2)

= 100 + (− 0.5)10 + 0.06 = 100.06 − 5 = 95.06