# Class 8 RD Sharma Solutions- Chapter 6 Algebraic Expressions And Identities – Exercise 6.6 | Set 1

**Question 1. Write the following squares of binomials as trinomials:**

**(i) (x + 2)**^{2}

^{2}

**Solution:**

x

^{2}+ 2 (x) (2) + 22x

^{2}+ 4x + 4

**(ii) (8a + 3b)**^{2}

^{2}

**Solution:**

(8a)

^{2}+ 2 (8a) (3b) + (3b)64a

^{2}+ 48ab + 9b^{2}

**(iii) (2m + 1)**^{2}

^{2}

**Solution:**

(2m)

^{2}+ 2 (2m) (1) + 1^{2}4m

^{2}+ 4m + 1

**(iv) (9a + 1/6)**^{2}

^{2}

**Solution:**

(9a)

^{2}+ 2 (9a) (1/6) + (1/6)^{2}81a

^{2}+ 3a + 1/36

**(v) (x + x**^{2}/2)^{2}

^{2}/2)

^{2}

**Solution:**

(x)

^{2}+ 2 (x) (x^{2}/2) + (x^{2}/2)^{2}x

^{2}+ x^{3}+ 1/4x^{4}

**(vi) (x/4 – y/3)**^{2}

^{2}

**Solution:**

(x/4)

^{2}– 2 (x/4) (y/3) + (y/3)^{2}1/16x

^{2}– xy/6 + 1/9y^{2}

**(vii) (3x – 1/3x)**^{2}

^{2}

**Solution:**

(3x)

^{2}– 2 (3x) (1/3x) + (1/3x)^{2}9x

^{2}– 2 + 1/9x^{2}

### (**viii) (x/y – y/x)**^{2}

^{2}

**Solution:**

(x/y)

^{2 }– 2 (x/y) (y/x) + (y/x)^{2}x

^{2}/y^{2}– 2 + y^{2}/x^{2}

**(ix) (3a/2 – 5b/4)**^{2}

^{2}

**Solution:**

(3a/2)

^{2}– 2 (3a/2) (5b/4) + (5b/4)^{2}9/4a

^{2 }– 15/4ab + 25/16b^{2}

**(x) (a**^{2}b – bc^{2})^{2}

^{2}b – bc

^{2})

^{2}

**Solution:**

(a

^{2}b)^{2}– 2 (a^{2}b) (bc^{2}) + (bc^{2})^{2}a

^{4}b^{4}– 2a^{2}b^{2}c^{2}+ b^{2}c^{4}

**(xi) (2a/3b + 2b/3a)**^{2}

^{2}

**Solution:**

(2a/3b)

^{2}+ 2 (2a/3b) (2b/3a) + (2b/3a)^{2}4a

^{2}/9b^{2}+ 8/9 + 4b^{2}/9a^{2}

**(xii) (x**^{2} – ay)^{2}

^{2}– ay)

^{2}

**Solution:**

(x

^{2})^{2}– 2 (x^{2}) (ay) + (ay)^{2}x

^{4}– 2x^{2}ay + a^{2}y^{2}

**Question 2. Find the product of the following binomials:**

**i) (2x + y) (2x + y)**

**Solution:**

2x (2x + y) + y (2x + y)

4x

^{2}+ 2xy + 2xy + y^{2}4x

^{2}+ 4xy + y^{2}

**(ii) (a + 2b) (a – 2b)**

**Solution:**

a (a – 2b) + 2b (a – 2b)

a

^{2}– 2ab + 2ab – 4b^{2}a

^{2}– 4b^{2}

**(iii) (a**^{2} + bc) (a^{2} – bc)

^{2}+ bc) (a

^{2}– bc)

**Solution:**

a

^{2}(a^{2}– bc) + bc (a^{2}– bc)a

^{4}– a^{2}bc + bca^{2}– b^{2}c^{2}a

^{4}– b^{2}c^{2}

**(iv) (4x/5 – 3y/4) (4x/5 + 3y/4)**

**Solution:**

4x/5 (4x/5 + 3y/4) – 3y/4 (4x/5 + 3y/4)

16/25x

^{2}+ 12/20yx – 12/20xy – 9y^{2}/1616/25x

^{2}– 9/16y^{2}

**(v) (2x + 3/y) (2x – 3/y)**

**Solution:**

2x (2x – 3/y) + 3/y (2x – 3/y)

4x

^{2}– 6x/y + 6x/y – 9/y^{2}4x

^{2}– 9/y^{2}

**(vi) (2a**^{3} + b^{3}) (2a^{3} – b^{3})

^{3}+ b

^{3}) (2a

^{3}– b

^{3})

**Solution:**

2a

^{3}(2a^{3}– b^{3}) + b^{3}(2a^{3}– b^{3})4a

^{6}– 2a^{3}b^{3}+ 2a^{3}b^{3}– b^{6}4a

^{6}– b^{6}

**(vii) (x**^{4} + 2/x^{2}) (x^{4} – 2/x^{2})

^{4}+ 2/x

^{2}) (x

^{4}– 2/x

^{2})

**Solution:**

= x

^{4}(x^{4}– 2/x^{2}) + 2/x^{2}(x^{4}– 2/x^{2})= x

^{8}– 2x^{2}+ 2x^{2}– 4/x^{4}= (x

^{8}– 4/x^{4})

**(viii) (x**^{3} + 1/x^{3}) (x^{3} – 1/x^{3})

^{3}+ 1/x

^{3}) (x

^{3}– 1/x

^{3})

**Solution:**

= x

^{3}(x^{3}– 1/x^{3}) + 1/x^{3}(x^{3}– 1/x^{3})= x

^{6}– 1 + 1 – 1/x^{6}= x

^{6}– 1/x^{6}

**Question 3. Using the formula for squaring a binomial, evaluate the following:**

**(i) (102)**^{2}

^{2}

**Solution:**

We can rewrite 102 as 100 + 2

(102)2 = (100 + 2)2

By simplification ,

(100 + 2)2 = (100)2 + 2 (100) (2) + 22

= 10000 + 400 + 4 = 10404

**(ii) (99)**^{2}

^{2}

**Solution:**

We can rewrite 99 as 100 – 1

(99)

^{2}= (100 – 1)^{2}On simplification,

(100 – 1)

^{2}= (100)^{2}– 2 (100) (1) + 1^{2}= 10000 – 200 + 1 = = 9801

**(iii) (1001)**^{2}

^{2}

**Solution:**

We can rewrite 1001 as 1000 + 1

(1001)

^{2}= (1000 + 1)^{2}On simplification ,

(1000 + 1)

^{2}= (1000)^{2}+ 2 (1000) (1) + 1^{2}= 1000000 + 2000 + 1 = 1002001

**(iv) (999)**^{2}

^{2}

**Solution:**

We can rewrite 999 as 1000 – 1

(999)

^{2}= (1000 – 1)^{2}By simplification,

(1000 – 1)

^{2}= (1000)2 – 2 (1000) (1) + 12= 1000000 – 2000 + 1 = 998001

**(v) (703)**^{2}

^{2}

**Solution:**

We can rewrite 700 as 700 + 3

(703)

^{2}= (700 + 3)^{2}By simplification,

(700 + 3)2 = (700)2 + 2 (700) (3) + 32

= 490000 + 4200 + 9 = 494209

**Question 4. Simplify the following using the formula: (a – b) (a + b) = a**^{2} – b^{2} :

^{2}– b

^{2}:

**(i) (82)**^{2} – (18)^{2}

^{2}– (18)

^{2}

**Solution:**

Here we will use the formula

(82)

^{2}– (18)^{2}= (82 – 18) (82 + 18)= 64 × 100

= 6400

**(ii) (467)**^{2} – (33)2^{2}

^{2}– (33)2

^{2}

**Solution:**

We will using the formula (a – b) (a + b) = a

^{2}– b^{2}(467)

^{2}– (33)^{2}= (467 – 33) (467 + 33)= (434) (500)

= 217000

**(iii) (79)**^{2} – (69)^{2}

^{2}– (69)

^{2}

**Solution:**

We will using the formula (a – b) (a + b) = a

^{2}– b^{2}(79)

^{2}– (69)^{2}= (79 + 69) (79 – 69)= (148) (10)

= 1480

**(iv) 197 × 203**

**Solution:**

We can rewrite 203 as 200 + 3 and 197 as 200 – 3

We will using the formula (a – b) (a + b) = a

^{2}– b^{2}197 × 203 = (200 – 3) (200 + 3)

= (200)

^{2}– (3)^{2}= 40000 – 9

= 39991

**(v) 113 × 87**

**Solution:**

We can rewrite 113 as 100 + 13 and 87 as 100 – 13

We can using the formula (a – b) (a + b) = a

^{2}– b^{2}113 × 87 = (100 – 13) (100 + 13)

= (100)

^{2}– (13)^{2}= 10000 – 169

= 9831

**(vi) 95 × 105**

**Solution:**

We can rewrite 95 as 100 – 5 and 105 as 100 + 5

We will using the formula (a – b) (a + b) = a

^{2}– b^{2}95 × 105 = (100 – 5) (100 + 5)

= (100)

^{2}– (5)^{2}= 10000 – 25

= 9975

**(vii) 1.8 × 2.2**

**Solution:**

We can rewrite 1.8 as 2 – 0.2 and 2.2 as 2 + 0.2

We will using the formula (a – b) (a + b) = a

^{2}– b^{2}1.8 × 2.2 = (2 – 0.2) ( 2 + 0.2)

= (2)

^{2}– (0.2)^{2}= 4 – 0.04

= 3.96

**(viii) 9.8 × 10.2**

**Solution:**

We can rewrite 9.8 as 10 – 0.2 and 10.2 as 10 + 0.2

We will using the formula (a – b) (a + b) = a

^{2}– b^{2}9.8 × 10.2 = (10 – 0.2) (10 + 0.2)

= (10)

^{2}– (0.2)^{2}= 100 – 0.04

= 99.96

**Question 5. Simplify the following using the identities:**

**(i) ((58)**^{2} – (42)^{2})/16

^{2}– (42)

^{2})/16

**Solution:**

We will using the formula (a – b) (a + b) = a

^{2}– b^{2}((58)

^{2}– (42)^{2})/16 = ((58-42) (58+42)/16)= ((16) (100)/16)

= 100

**(ii) 178 × 178 – 22 × 22**

**Solution:**

We will using the formula (a – b) (a + b) = a

^{2}– b^{2}178 × 178 – 22 × 22 = (178)

^{2}– (22)^{2}= (178-22) (178+22)

= 200 × 156

= 31200

**(iii) (198 × 198 – 102 × 102)/96**

**Solution:**

We using the formula (a – b) (a + b) = a

^{2}– b^{2}(198 × 198 – 102 × 102)/96 = ((198)2 – (102)2)/96

= ((198-102) (198+102))/96

= (96 × 300)/96

= 300

**(iv) 1.73 × 1.73 – 0.27 × 0.27**

**Solution:**

We will using the formula (a – b) (a + b) = a

^{2}– b^{2}1.73 × 1.73 – 0.27 × 0.27 = (1.73)2 – (0.27)2

= (1.73-0.27) (1.73+0.27)

= 1.46 × 2

= 2.92

**(v) (8.63 × 8.63 – 1.37 × 1.37)/0.726**

**Solution:**

We will using the formula (a – b) (a + b) = a

^{2}– b^{2}(8.63 × 8.63 – 1.37 × 1.37)/0.726 = ((8.63)2 – (1.37)2)/0.726

= ((8.63-1.37) (8.63+1.37))/0.726

= (7.26 × 10)/0.726

= 72.6/0.726

= 100

**Question 6. Find the value of x, if:**

**(i) 4x = (52)**^{2} – (48)^{2}

^{2}– (48)

^{2}

**Solution:**

We will using the formula (a – b) (a + b) = a

^{2}– b^{2}4x = (52)

^{2}– (48)^{2}4x = (52 – 48) (52 + 48)

4x = 4 × 100

4x = 400

x = 100

**(ii) 14x = (47)**^{2} – (33)^{2}

^{2}– (33)

^{2}

**Solution:**

We will using the formula (a – b) (a + b) = a2 – b2

14x = (47)2 – (33)2

14x = (47 – 33) (47 + 33)

14x = 14 × 80

x = 80

**(iii) 5x = (50)**^{2} – (40)^{2}

^{2}– (40)

^{2}

**Solution:**

We using the formula (a – b) (a + b) = a2 – b2

5x = (50)

^{2}– (40)^{2}5x = (50 – 40) (50 + 40)

5x = 10 × 90

5x = 900

x = 180

**Question 7. If x + 1/x =20, find the value of x**^{2} + 1/ x^{2}.

^{2}+ 1/ x

^{2}.

**Solution:**

Given equation in the x + 1/x = 20

when squaring both sides, we get

(x + 1/x)

^{2}= (20)^{2}x

^{2}+ 2 × x × 1/x + (1/x)^{2}= 400x

^{2}+ 2 + 1/x^{2}= 400x

^{2}+ 1/x^{2}= 398

**Question 8. If x – 1/x = 3, find the values of x**^{2} + 1/ x^{2} and x^{4} + 1/ x^{4}.

^{2}+ 1/ x

^{2}and x

^{4}+ 1/ x

^{4}.

**Solution:**

Given in the question x – 1/x = 3

when squaring both sides,

(x – 1/x)

^{2}= (3)^{2}x

^{2}– 2 × x × 1/x + (1/x)^{2}= 9x

^{2}– 2 + 1/x^{2}= 9x

^{2}+ 1/x^{2}= 9+2x

^{2}+ 1/x^{2}= 11Now again when we square on both sides ,

(x

^{2}+ 1/x^{2})^{2}= (11)^{2}x

^{4}+ 2 × x^{2}× 1/x^{2}+ (1/x^{2})^{2}= 121x

^{4 }+ 2 + 1/x^{4}= 121x

^{4}+ 1/x^{4}= 121-2x

^{4 }+ 1/x^{4}= 119x

^{2}+ 1/x^{2}= 11x

^{4}+ 1/x^{4}= 119

**Question 9. If x**^{2} + 1/x^{2} = 18, find the values of x + 1/ x and x – 1/ x.

^{2}+ 1/x

^{2}= 18, find the values of x + 1/ x and x – 1/ x.

**Solution:**

Given in the question x

^{2}+ 1/x^{2}= 18When adding 2 on both sides,

x

^{2}+ 1/x^{2}+ 2 = 18 + 2x

^{2}+ 1/x^{2}+ 2 × x × 1/x = 20(x + 1/x)

^{2}= 20x + 1/x = √20

When subtracting 2 from both sides,

x

^{2}+ 1/x^{2}– 2 × x × 1/x = 18 – 2(x – 1/x)

^{2}= 16x – 1/x = √16

x – 1/x = 4

**Question 10. If x + y = 4 and xy = 2, find the value of x**^{2} + y^{2}

^{2}+ y

^{2}

**Solution:**

We know that x + y = 4 and xy = 2

Upon squaring on both sides of the given expression, we get

(x + y)

^{2}= 42x

^{2}+ y^{2}+ 2xy = 16x

^{2}+ y^{2}+ 2 (2) = 16 (since x y=2)x

^{2}+ y^{2}+ 4 = 16x

^{2}+ y^{2}= 16 – 4x

^{2}+ y^{2}=12