Exercise 2.5

1. Use suitable identities to find the following products

(i) (x + 5)(x + 2)

Solution:

x2 + + + 10

= x2 + + 10

(ii) (x - 5)(x - 5)

Solution:

x−52 = x2 - 2(x1)() + 52

= x2 - x +

(iii) (3x + 2)(3x - 2)

Solution:

3x2 - 22 = 9x23x2 -

(iv) x2+1xx2−1x

Solution:

x22 - 1x2 = x24x22 - 1x2

(v) (1 + x)(1 + x)

Solution:

1+x2 = + 2x3x + x2

2. Evaluate the following products without actual multiplication.

(i) 101 × 99

Solution:

(100 + 1)(100 - 1) = 1002 - 12

= -

=

(ii) 999 × 999

Solution:

1000−12 = 10002 - 2()(1) + 12

= - + 1

=

(iii) 50 12 × 49 12

Solution:

(50 + 12)(50 - 12) = 502 - (12)²

= -

= 1000−14 =

(iv) 501 × 501

Solution:

500+12 = 50025012 + 2(500501)() + 121

= + + 1

=

(v) 30.5 × 29.5

Solution:

(30 + 0.5)(30 - 0.5) = 302 - 0.52

= -

=

3. Factorise the following using appropriate identities.

(i) 16x2 + 24xy + 9y2

Solution:

Given 16x2 + 24xy + 9y2

(4x24x) + 2()() + (3y23y)

= 4x+3y24x−3y2

(ii) 4y2 - 4y + 1

Solution:

Given 4y2 - 4y + 1

(2y2y2) - 2()() + 121

= 2y−122y+12

(iii) x225 - y24

Solution:

Given x225 - y24

(x52x252) -(y22y42)

= x5−y2x5+y2x5+y22

(iv) 18a2 - 50

Solution:

Given 18a2 - 50

(9a2 - 25)

= 2((3a23a) - 525)

= 2(3a-53a+5)(3a + 5)

(v) x2 + 5x + 6

Solution:

Given x2 + 5x + 6

x2 + + 3x + 6

= x (x+2x-2) + (x+2)

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

(vi) 3p2 - 24p + 36

Solution:

Given 3p2 - 24p + 36

(p2 - 8p + 12)

= 3(p2 - 2p - + 12)

=3(p(p-2p-2p) - 6(p-2p+2))

= 3(p-6)(p-2)(p-6)(p+2)

4. Expand each of the following, using suitable identities

(i) x+2y+4z2

Solution:

Given x+2y+4z2

x2 + 2y2 + 4z2 + 2()(2y) + 2()(4z) + 2(x)()

= x2 + 4y22y2 + 16z2 + + 16yz + z

(ii) 2a−3b3

Solution:

Given 2a−3b3

2a3 - 3(2a2a2)(3b) + 3(2a)(3b2b2) - 3b3

= 8a38a3 - a2b + ab2 - b3

(iii) −2a+5b−3c2

Solution:

Given −2a+5b−3c2

−2a2 + 5b2 + −3c2 + 2()(5b) + 2()(-3c) + 2(-2a)()

= a2 + b2 + c2 - - 30bc +

(iv) a4−b+22

Solution:

Given a4−b+22

a42 + −b2 + 22 + 2(a4a2)(-b) + 2(-b)() + 2a42

= a216a28 + b2 + - ab2 - +

(v) p+13

Solution:

Given p+13

p3 + 3p23p (1) + 3p121 + 13

= p3 + 3p2p2 + + 1

(vi) x3−y23

Solution:

Given x3−y23

x33 - 3x32(y2y4) + 3(x3x6)y22 - y23

= x327x39 - x2y6xy26 + xy24x2y4 + −y38y38

5. Factorise

(i) 25x2 + 16y2 + 4z2 - 40xy + 16yz - 20xz

Solution:

5x215x2 + 4y24y2 + 2z22z2 + 2()(-4y) + 2(-4y)() + 2()(-2z)

= 5x−4y−2z25x+4y−2z2

(ii) 9a2+4b2+16c2+12ab−16bc−24ca

Solution:

(3a23a) + (2b22b) + (4c24c) + 2()(2b) + 2()(-4c) + 2(3a)()

= 3a+2b−4c23a−2b−4c2

6. If a + b + c = 9 and ab + bc + ca = 26, then find a2 + b2 + c2.

Solution:

We know that a+b+c2 = a2 + b2 + c2 + (ab + bc + ca).

Substituting the given values, we have 9232 = a2 + b2 + c2 + 2().

= a2 + b2 + c2 + .

Therefore, a2 + b2 + c2 = - = .

7. Evaluate the following using suitable identities.

(i) 993

Solution:

100−13 = 1003 - 3(1002102)(1) + 3(100)(121) - 13

= - + -

=

(ii) 1023

Solution:

100+23 = 1003 + 3(1002)(2) + 3(100)(22) + 23

= + + +

=

(iii) 9983

Solution:

1000−23 = 10003 - 3(10002)(2) + 3(1000)(22) - 23

= - + -

=

(iv) 10013

Solution:

1000+13 = 10003 + 3(10002)(1) + 3(1000)(12) + 13

= + + +

= 10030030011000303001

8. Factorise each of the following

(i) 8a3 + b3 + 12a2b + 6ab2

Solution:

Given 8a3 + b3 + 12a2b + 6ab2

2a3 + b3 + 3(2a2a2)b + 3()b2

= 2a+b32a−b3

(ii) 8a3 - b3 - 12a2b + 6ab2

Solution:

Given 8a3 - b3 - 12a2b + 6ab2

(2a32a2) - b3 - 32a2b + 3(2a)b2

= 2a−b3−2a−b3

(iii) 1 - 64a3 - 12a + 48a2

Solution:

Given 1 - 64a3 - 12a + 48a2

13 - (4a34a3) - 3(12)(4a) + 3(1)(4a2)

= 1−4a31+4a3

(iv) (125/8)p3 - (75/4)p2 + (15/2)p - 1

Solution:

Given (125/8)p3 - (75/4)p2 + (15/2)p - 1

(5p235p23) - 35p22(1) + 3(5p/2)12 - 13

= 5p2−135p2+13

9. Verify (i) x3 + y3 = (x + y)(x2 - xy + y2) (ii) x3 - y3 = (x - y)(x2 + xy + y2) using some non-zero positive integers and check by actual multiplication. Can you call these as identities ?

Solution:

(i) Let x = 2 and y = 3.

LHS = 2322 + 3332 = + =

RHS = (2 + 3)(22 - 2×3 + 32) = ( - + ) = × = .

LHS =≠ RHS.

(ii) Let x = 3 and y = 2.

LHS = 3332 - 2322 = - =

RHS = (3 - 2)(32 + 3×2 + 22) = ( + + ) = .

LHS =≠ RHS.

Yes, these can be called identities because they hold true for all values of x and y.

10. Factorise (i) 27a3 + 64b3 (ii) 343y3 - 1000 using the above results.

Solution:

(i) 27a3 + 64b3

3a33a2 + 4b34b2

Since we know: a3 + b3 = (a + b)(a2 - ab + b2)

= ( + )(3a23a3 - ()() + 4b24b3)

= (3a + 4b)(9a29a3 - ab + 16b24b2)

(ii) 343y3 - 1000

7y37y2 - 103102

a3 - b3 = (a - b)(a2 + ab + b2)

= ( - )(7y27y3 + (7yy)() + 102103)

= (7y - 10)(49y245y2 + y + )

11. Factorise 27x3 + y3 + z3 - 9xyz using identity.

Solution:

Given 27x3 + y3 + z3 - 9xyz

3x33x2 + y3 + z3 - 3()(y)(z)

a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)

= ( + + ) ( 3x23x3 + y2y3 + z2z3 - ()y - z - (3x))

= (3x + y + z)( x2 + y2 + z2 - - - )

12. Verify that x3 + y3 + z3 - 3xyz = (1/2)(x + y + z)[x−y2 + y−z2 + z−x2]

Solution:

We will start with the right-hand side (RHS) and show that it simplifies to the left-hand side (LHS).

RHS = 12x+y+zx−y2+y−z2+z−x2

= 12 (x + y + z)((x22x - 2xyxy + y22y) + (y22y - 2yzyz + z22z) + z22z - 2zxzx + z22x)

= 12 (x + y + z) ( x2 + y2 + z2 - 2xy - 2yz - 2zx)

= 12 (x + y + z)( (x2 + y2 + z2 - xy - yz - zx))

= (x + y + z)(x2 + y2 + z2 - xy - yz - zx)

= x(x2 + y2 + z2 - xy - yz - zx) + y(x2 + y2 + z2 - xy - yz - zx) + z(x2 + y2 + z2 - xy - yz - zx) (Distributing)

= x3 + xy2 + xz2 - x2y - xyz - x2z + yx2 + y3 + yz2 - xy2 -y2z - xyz + zx2 + zy2 + z3 - zxy - yz2 - xz2 (Expanding)

= x32x3 + y32xy2 + z32z3 +

Since the RHS simplifies to x3+y3+z3−3xyz, which is the LHS, the identity is verified.

13. (a) If x + y + z = 0, show that x3 + y3 + z3 = 3xyz.

Solution:

Given: x + y + z = 0

To Prove: x3 + y3 + z3 = 3xyz

Proof:

From the given condition, we have: x + y =

Cube both sides of the equation: x+y3 = −z3, which expands to x3x2 + 3x22xy + 3xy2y + y3y2 = −z3z3.

Rearrange the equation to isolate x3 + y3 + z3: x3 + y3 + z3 = −3x2y3x2y + −3xy23xy2

Factor out -3xy from the right-hand side: x3 + y3 + z3 = -3xy( x+y2x+y ).

Substitute x + y = -z (from the given condition) into the equation: x3 + y3 + z3 = -3xy(-z)

Simplify: x3 + y3 + z3 =

Therefore, if x + y + z = 0, then x3 + y3 + z3 = 3xyz.

(b) Show that a−b3 + b−c3 + c−a3 = 3 (a – b) (b – c) (c – a)

Solution:

Let x = a - b, y = b - c, and z = c - a.

Then, x + y + z = (a - b) + (b - c) + (c - a) = a - b + b - c + c - a = .

From part (a), we know that if x + y + z = 0, then x3 + y3 + z3 = .

Therefore, we have a−b3 + b−c3 + c−a3 = (a - b)(b - c)(c - a).

Substituting x, y, and z back into the equation, we get:

a−b3 + b−c3 + c−a3 = 3(a - b)(b - c)(c - a).

Hence, a−b3 + b−c3 + c−a3 = 3 (a – b) (b – c) (c – a) is proven.

14. Without actual calculating the cubes, find the value of each of the following:

Solution:

(i) −103 + 73 + 33

Solution:

Let a = , b = , and c = .

Check if a + b + c = 0: -10 + 7 + 3 = .

Since a + b + c = 0, we can use the identity a3 + b3 + c3 = 3abc.

Therefore, −103 + 73 + 33 = 3(-10)(7)(3).

Calculate the product: 3(-10)(7)(3) = .

Thus, −103 + 73 + 33 = -630.

(ii) (28)³ + (−15)³ + (−13)³

Solution:

Let a = , b = , and c = .

Check if a + b + c = 0: 28 + (-15) + (-13) = 28 - 15 - 13 = .

Since a + b + c = 0, we can use the identity a3 + b3 + c3 = 3abc.

Therefore, 283 + −153 + −133 = 3(28)(-15)(-13).

Calculate the product: 3(28)(-15)(-13) = .

Thus, 283 + −153 + −133 = 16380.

15. Give possible expressions for the length and breadth of the rectangle whose area is given by

(i) 4a2 + 4a - 3

Solution:

We need to factor the quadratic expression 4a2 + 4a - 3.

We look for two numbers whose product is (4)(-3) = and whose sum is .

The numbers are 6 and .

Rewrite the middle term: 4a2 + 6a - 2a - 3.

Factor by grouping: (2a + 3) - (2a + 3).

Factor out the common term: (2a + 3)(2a - 1)(2a + 3)(2a + 1).

Possible expressions for length and breadth: (2a + 3) and (2a - 1).

(ii) 25a2 - 35a + 12

Solution:

We need to factor the quadratic expression 25a2 - 35a + 12.

We look for two numbers whose product is (25)(12) = and whose sum is .

The numbers are and .

Rewrite the middle term: 25a2 - 20a - 15a + 12.

Factor by grouping: (5a - 4) - (5a - 4).

Factor out the common term: (5a - 4)(5a - 3)(5a - 4)(5a + 3)

Possible expressions for length and breadth: (5a - 4) and (5a - 3).

16. What are the possible polynomial expressions for the dimensions of the cuboids whose volumes are given below?

(i) 3x3 − 12x

Solution:

Factor out the common factor 3x: x2−4.

Factor the difference of squares: 3x(x + 2)(x-2x+2).

Possible dimensions: 3x, (x + 2), (x - 2).

(ii) 12y2 + 8y − 20

Solution:

Factor out the common factor 4: (3y22y2 + y - ).

Factor the quadratic expression 3y2 + 2y - 5.

Look for two numbers whose product is (3)(-5) = and whose sum is 2.

The numbers are 5 and -3.

Rewrite the middle term: 3y2 + - - 5.

Factor by grouping: (3y + 5) - (3y + 5).

Factor out the common term: (3y + 5)(y - 1).

Therefore, 12y2 + 8y - 20 = 4(3y + 5)(y - 1)3(3y - 5)(y - 1).

Possible dimensions: 4, (3y + 5), (y - 1).

17. If 2a2+b2 = a+b2, then show that a = b.

Solution:

Given: 2a2+b2 = a+b2

Expand the right side: 2a2+b2 = a22a + 2abab + b2b.

Distribute the 2 on the left side: 2a2 + 2b2 = a2 + 2ab + b2.

Move all terms to the left side: 2a2 + 2b2 - a2 - 2ab - b2 = 0.

Simplify: a22a2 + -2ab2ab + b22b2 = 0.

Recognize the left side as a perfect square: a−b2a+b2 = 0.

Take the square root of both sides: a-ba+b = 0.

Solve for a: a = .

Therefore, a = b is proven.