NCERT Class 9 Mathematics Textbook for Blind and Visually Impaired Students made Screen Readable by Professor T K bansal.

From our earlier classes, we may recall that an algebraic identity is an algebraic equation that is true for all values of the variables occurring in it. We have studied the following algebraic identities in earlier classes:

Identity 1 :
\[(x\ +\ y)^2\ =\ x^2\ +\ 2xy\ +\ y^2\]

Identity 2 :
\[(x\ −\ y)^2\ =\ x^2\ −\ 2xy\ +\ y^2\]

Identity 3 :
\[x^2\ −\ y^2\ =\ (x+y)\ ×\ (x−y)\]

Identity 4 :
\[(x + a)\ ×\ (x + b)\ =\ x^2\ +\ (a+b)x\ +\ ab\]

We have been using some of these algebraic identities to factorise the algebraic expressions. We can also see their utility in computations.

Example 16

Find the following products using appropriate identities:
\[(i)\ (x + 3)\ (x + 3)\]
\[(ii)\ (x − 3)\ (x + 5)\]

Solution :

(i) Here we can use Identity 1 :
\[(x + y)^2\ =\ x^2\ +\ 2xy\ +\ y^2.\]
Putting y = 3 in it,
\[(x + 3) (x + 3)\ =\ (x + 3)^2\ =\ x^2\ +\ 2(x)(3)\ +\ (3)^2\]
\[=\ x^2\ +\ 6x\ +\ 9\]

(ii) Using Identity 4 above, i.e.,
\[(x + a)\ (x + b)\ =\ x^2\ +\ (a + b)x\ +\ ab,\]
we have
\[(x − 3)\ (x + 5)\ =\ x^2\ +\ (−3 + 5)x\ +\ (−3)(5)\]
\[=\ x^2\ +\ 2x\ −\ 15\]

Example 17

Evaluate 105 × 106 without multiplying directly.

Solution :

105 × 106 = (100 + 5) × (100 + 6)
= (100)^2 + (5 + 6) (100) + (5 × 6), using Identity 4
= 10000 + 1100 + 30
= 11130

We have seen some uses of the identities listed above in finding the product of some given expressions. These identities are useful in factorisation of algebraic expressions also, as we will see in the following examples.

Example 18

Factorise:
\[(i)\ 49a^2\ +\ 70ab\ +\ 25b^2\]
\[(ii)\ \frac{25}{4}×x^2\ −\ \frac{y^2}{9}\]

Solution :

(i) Here you can see that
49a^2 = (7a)^2,
25b^2 = (5b)^2,
70ab = 2(7a) (5b)
Comparing the given expression with x^2 + 2xy + y^2, we observe that x = 7a and y = 5b.

Using Identity 1, we get
49a^2 + 70ab + 25b^2 = (7a + 5b)^2 = (7a + 5b) (7a + 5b)

(ii) We have
(25/4)×x^2 − (y^2)/9 = (5/2)x^2 − (y/3)^2
Now comparing it with Identity 3, we get
(25/4) x^2 − y^2/9 = (5/2)x^2 − (y/3)^2 = ((5/2)x + y/3) ((5/2)x − y/3)

So far, all our identities involved products of binomials. Let us now extend the Identity I to a trinomial x + y + z. We shall compute (x + y + z)^2 by using Identity 1.

Let x + y = t. Then,
(x + y + z)^2 = (t + z)^2
= t^2 + 2tz + z^2 (Using Identity 1)
= (x + y)^2 + 2(x + y)z + z^2 (Substituting the value of t)
= x^2 + 2 x y + y^2 + 2xz + 2yz + z^2 (Using Identity 1)
= x^2 + y^2 + z^2 + 2 x y + 2 y z + 2 z x (Rearranging the terms)

So, we get the following identity:

Identity 5 :
\[(x\ +\ y\ +\ z)^2\ =\ x^2\ +\ y^2\ +\ z^2\ +\ 2xy\ +\ 2yz\ +\ 2zx\]

Remark :We call the right hand side expression the expanded form of the left hand side expression.
Note that the expansion of (x + y + z)^2 consists of three square terms and three product terms.

Example 19

Write (3a + 4b + 5c)^2 in expanded form.

Solution :

Comparing the given expression with (x + y + z)^2, we find that
x = 3a,
y = 4b and
z = 5c.

Therefore, using Identity 5, we have
(3a + 4b + 5c)^2 = (3a)^2 + (4b)^2 + (5c)^2 + 2(3a)(4b) + 2(4b)(5c) + 2(5c)(3a)
= 9a^2 + 16b^2 + 25c^2 + 24ab + 40bc + 30ac

Example 20

Expand (4a − 2b − 3c)^2.

Solution :

Using Identity 5, we have
(4a − 2b − 3c)^2 = [4a + (−2b) + (−3c)]^2
= (4a)^2 + (−2b)^2 + (−3c)^2 + 2(4a)(−2b) + 2(−2b)(−3c) + 2(−3c)(4a)
= 16a^2 + 4b^2 + 9c^2 − 16ab + 12bc − 24ac

Example 21

Factorise 4x^2 + y^2 + z^2 − 4x y − 2y z + 4xz.

Solution :

We have 4x^2 + y^2 + z^2 − 4x y − 2y z + 4xz = (2x)^2 + (−y)^2 + (z)^2 + 2(2x) × (−y)+ 2(−y) × (z) + 2(2x) × (z) = [2x + (−y) + z]^2 (Using Identity V)
= (2x − y + z)^2 = (2x − y + z)(2x − y + z)

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So far, we have dealt with identities involving second degree terms. Now let us extend Identity 1 to compute (x + y)^3 . We have:

\[(x\ +\ y)^3\ =\ (x\ +\ y)\ ×\ (x\ +\ y)^2\]
\[=\ (x\ +\ y)\ ×\ (x^2\ +\ 2xy\ +\ y^2)\]
\[=\ x(x^2\ +\ 2xy\ +\ y^2)\ +\ y(x^2\ +\ 2xy\ +\ y^2)\]
\[=\ x^3\ +\ 2x^2y\ +\ xy^2\ +\ x^2y\ +\ 2xy^2\ +\ y^3\]
\[=\ x^3\ +\ 3x^2y\ +\ 3xy^2\ +\ y^3\]
\[=\ x^3\ +\ y^3\ +\ 3xy(x\ +\ y)\]

So, we get the following identity:

Identity 6 :
\[(x\ +\ y)^3\ =\ x^3\ +\ y^3\ +\ 3xy(x\ +\ y)\]

Also, by replacing y by − y in the Identity 6, we get

Identity 7 :
\[(x\ −\ y)^3\ =\ x^3\ −\ y^3\ −\ 3xy(x\ −\ y)\]
\[=\ x^3\ −\ 3x^2y\ +\ 3xy^2\ −\ y^3\]

Example 22

Write the following cubes in the expanded form:
\[(i)\ (3a\ +\ 4b)^3\]
\[(ii)\ (5p\ −\ 3q)^3\]

Solution :

(i) Comparing the given expression with (x + y)^3, we find that
x = 3a and y = 4b.
So, using Identity VI, we have:
(3a + 4b)^3 = (3a)^3 + (4b)^3 + 3(3a)(4b)(3a + 4b)
= 27a^3 + 64b^3 + 108a^2b + 144ab^2

(ii) Comparing the given expression with (x − y)^3, we find that
x = 5p, y = 3q.
So, using Identity VII, we have:
(5p − 3q)^3 = (5p)^3 − (3q)^3 − 3(5p)(3q)(5p − 3q)
= 125p^3 − 27q^3 − 225p^2q + 135pq^2

Example 23

Evaluate each of the following using suitable identities:
\[(i)\ (104)^3\]
\[(ii)\ (999)^3\]

Solution :

(i) We have
(104)^3 = (100 + 4)^3
= (100)^3 + (4)^3 + 3(100)(4)(100 + 4) (Using Identity VI)
= 1000000 + 64 + 124800
= 1124864

(ii) We have
(999)^3 = (1000 − 1)^3
= (1000)^3 − (1)^3 − 3(1000)(1)(1000 − 1) (Using Identity VII)
= 1000000000 − 1 − 2997000
= 997002999

Example 24

Factorise
\[8x^3\ +\ 27y^3\ +\ 36x^2y\ +\ 54xy^2\]

Solution :

The given expression can be written as
(2x)^3 + (3y)^3 + 3(4x^2)(3y) + 3(2x)(9y^2)
= (2x)^3 + (3y)^3 + 3(2x)^2(3y) + 3(2x)(3y)^2
= (2x + 3y)^3 (Using Identity VI)
= (2x + 3y)(2x + 3y)(2x + 3y)

Now consider (x + y + z) × (x^2 + y^2 + z^2 − x y − y z − zx)
On expanding, we get the product as
x(x^2 + y^2 + z^2 − x y − y z − zx) + y(x^2 + y^2 + z^2 − x y − y z − zx)+ z(x^2 + y^2 + z^2 − x y − y z − zx)
= x^3 + x y^2 + xz^2 − x^2y − x y z − zx^2 + x^2y+ y^3 + y z^2 + x y^2 − y^2z − x y z + x^2z + y^2z + z^3 – x y z − y z^2 − xz^2
= x^3 + y^3 + z^3 − 3x y z (On simplification)
So, we obtain the following identity:

Identity 8 :
\[x^3\ +\ y^3\ +\ z^3\ −\ 3xyz\ =\ (x\ +\ y\ +\ z)\ ×\ (x^2\ +\ y^2\ +\ z^2\ −\ xy\ −\ yz\ −\ zx)\]

Example 25

Factorise :
\[8x^3\ +\ y^3\ +\ 27z^3\ −\ 18xyz\]

Solution :

Here, we have
8x^3 + y^3 + 27z^3 − 18x y z
= (2x)^3 + y^3 + (3z)^3 − 3(2x)(y)(3z)
= (2x + y + 3z) × [(2x)^2 + y^2 + (3z)^2 − (2x)(y) − (y)(3z) − (2x)(3z)]
= (2x + y + 3z) × (4x^2 + y^2 + 9z^2 − 2x y − 3y z − 6xz)

EXERCISE 2.5

Q1. Use suitable identities to find the following products:
\[(i)\ (x\ +\ 4)\ (x\ +\ 10)\]
\[(ii)\ (x\ +\ 8)\ (x\ −\ 10)\]
\[(iii)\ (3x\ +\ 4)\ (3x\ −\ 5)\]
\[(iv)\ (y^2\ +\ 3/2)\ (y^2\ −\ 3/2)\]
\[(v)\ (3\ −\ 2x)\ (3\ +\ 2x)\]

A1.
(i) x^2 + 14x + 40
(ii) x^2 − 2x − 80
(iii) 9x^2 − 3x − 20
(iv) y^4 − (9/4)
(v) 9 − 4x^2

Q2. Evaluate the following products without multiplying directly:
(i) 103 × 107
(ii) 95 × 96
(iii) 104 × 96

A2.
(i) 11021
(ii) 9120
(iii) 9984

Q3. Factorise the following using appropriate identities:
(i) 9x^2 + 6x y + y^2
(ii) 4y^2 − 4y + 1
(iii) x^2 − (y^2)/100

A3.
(i) (3x + y) (3x + y)
(ii) (2y − 1) (2y − 1)
(iii) (x + (y/10) (x − (y/10)

Q4. Expand each of the following, using suitable identities:
(i) (x + 2y + 4z)^2
(ii) (2x − y + z)^2
(iii) (−2x + 3y + 2z)^2
(iv) (3a − 7b − c)^2
(v) (−2x + 5y − 3z)^2
(vi) [(1/4) a − (1/2) b +1]^2

A4.
(i) x^2 + 4y^2 + 16z^2 + 4x y + 16y z + 8xz
(ii) 4x^2 + y^2 + z^2 − 4x y − 2y z + 4xz
(iii) 4x^2 + 9y^2 + 4z^2 − 12x y + 12y z − 8xz
(iv) 9a^2 + 49b^2 + c^2 −42a b + 14bc − 6a c
(v) 4x^2 + 25y^2 + 9z^2 − 20x y − 30y z + 12xz
(vi) (a^2/16) + (b^2/4) + 1 − (a b/4) − b + (a/2)

Q5. Factorise:
(i) 4x^2 + 9y^2 + 16z^2 + 12x y − 24y z − 16x z
(ii) 2x^2 + y^2 + 8z^2 − 2 (√2) x y + 4 (√2) y z − 8x z

A5. (i) (2x + 3y − 4z) (2x + 3y − 4z)
(ii) (−√2 x + y + 2 √2) (−√2 x + y + 2 √2z)

Q6. Write the following cubes in expanded form:
(i) (2x + 1)^3
(ii) (2a − 3b)^3
(iii) [(3/2) x + 1]^3
(iv) [x−(2/3)y]^3

A6.
(i) 8x^3 + 12x^2 + 6x + 1
(ii) 8a^3 − 27b^3 − 36a^2b + 54a b^2
(iii) (27/8) x^3 + (27/4) x^2 + (9/2)x + 1
(iv) x^3 − (8/27) y^3 + 2x^2y + 4x y^2 / 3

Q7. Evaluate the following using suitable identities:
(i) (99)^3
(ii) (102)^3
(iii) (998)^3

A7.
(i) 970299
(ii) 1061208
(iii) 994011992

Q8. Factorise each of the following:
(i) 8a^3 + b^3 + 12a^2b + 6ab^2
(ii) 8a^3 − b^3 − 12a^2b + 6ab^2
(iii) 27 − 125a^3 − 135a + 225a^2
(iv) 64a^3 − 27b^3 − 144a^2b + 108ab^2
(v) 27p^3 − 1/216 − (9/ 2) p^2 +(1/4) p

A8.
(i) (2a + b) × (2a + b) × (2a + b)
(ii) (2a − b) × (2a − b) × (2a − b)
(iii) (3 − 5a) × (3 − 5a) × (3 − 5a)
(iv) (4a − 3b) × (4a − 3b) × (4a − 3b)
(v) (3p − 1/6) × (3p − 1/6) × (3p − 1/6)

Q9. Verify :
(i) x^3 + y^3 = (x + y) × (x^2 – x y + y^2)
(ii) x^3 − y^3 = (x − y) (x^2 + xy + y^2)

Q10. Factorise each of the following:
(i) 27y^3 + 125z^3
(ii) 64m^3 − 343n^3
[Hint : See Question 9.]

A10.
(i) (3y + 5z) × (9y^2 + 25z^2 − 15y z)
(ii) (4m − 7n) × (16m^2 + 49n^2 + 28mn)

Q11. Factorise : 27x^3 + y^3 + z^3 − 9x y z

A11. (3x + y + z) × (9x^2 + y^2 + z^2 − 3x y – y z − 3xz)

Q12. Verify that x^3 + y^3 + z^3 − 3x y z = (½) (x + y + z) × [( x − y)^2 + ( y − z )^2+ (z − x)^2]

A12. Simplify RHS.

Q13. If x + y + z = 0, show that x^3 + y^3 + z^3 = 3x y z.

A13. Put x + y + z = 0 in Identity VIII.

Q14. Without actually calculating the cubes, find the value of each of the following:
(i) (−12)^3 + (7)^3 + (5)^3
(ii) (28)^3 + (−15)^3 + (−13)^3

A14.
(i) −1260. Let a = −12, b = 7, c = 5. Here a + b + c = 0. Use the result given in Q13.
(ii) 16380

Q15. Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:
(i) Area : 25a^2 − 35a + 12
(ii) Area : 35y^2 + 13y −12

A15.
(i) One possible answer is : Length = 5a − 3, Breadth = 5a − 4
(ii) One possible answer is : Length = 7y − 3, Breadth = 5y + 4

Q16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
(i) Volume : 3x^2 − 12x
(ii) Volume : 12k y^2 + 8k y − 20k

A16.
(i) One possible answer is : 3, x and x − 4.
(ii) One possible answer is : 4k, 3y + 5 and y − 1.