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# Algebraic Identities

An identity is an equality that remains true regardless of the values chosen for its variables.

Algebraic Identities

An identity is an equality that remains true regardless of the values chosen for its variables.

1. (a + b )2 a 2 + 2ab +b2

2. (a b )2 a 22ab +b2

3. (a + b )(a b) a 2 b2

4. (x + a )(x +b) x 2 +(a + b )x +ab

Note

(i) a2 + b2 = (a +b)2 −2ab

(ii) a2 + b2 = (a − b )2 + 2ab

Example 3.16

Expand the following using identities: (i) (3x + 4y)2 (ii) (2a − 3b)2 (iii) (5x + 4y)(5x 4y) (iv) (m + 5)(m 8)

Solution

(i) (3x +  4y)2           [we have (a + b)2  = a2 + 2ab + b2 ]

(3x + 4y)2 = (3x)2 + 2(3x)(4y) +(4y)2  [put [a = 3x,  b = 4y]

= 9x2 + 24xy +16y2

(ii) (2a − 3b)2

[ we have (a − b)2 = a2 + 2ab + b2 ]

(2a − 3b)2 = (2a)2 −2(2a)(3b) +(3b)2    put [a = 2a, b = 3b]

= 4a 212ab + 9b2

(iii) (5x + 4y)(5x − 4y)

[ we have (a + b)(a −b) = a2b2]

(5x + 4y)(5x − 4y) = (5x)2 −(4y)2 put [a = 5x b = 4y]

= 25x2 −16y2

(iv) (m + 5) (m − 8)

[we have (x + a)(x-b) = x2 + (a-b)x-ab]

(m + 5) (m − 8) = m2 +(5 − 8)m −(5)(8)   put [x = m, a = 5, b = 8]

= m2 − 3m – 40

## 1. Expansion of Trinomial (a+b +c)2

We know that (x + y)2 = x 2 + 2xy + y2

Put x = a +b, y = c

Then, (a + b +c)2 = (a +b )2 + 2(a +b)(c ) +c2

= a 2 + 2ab + b2 + 2ac + 2bc +c2

= a2 +b2 + c 2 + 2ab + 2bc + 2ca

Thus, (a + b +c )2 a 2 +b 2 + c 2 + 2ab + 2bc + 2ca

Example 3.17

Expand (a b +c)2

Solution

Replacing ‘b’ by ‘-b ’ in the expansion of

(a + b +c)2 = a 2 +b2 + c 2 + 2ab + 2bc + 2ca

(a + (b) +c)2 = a 2 +(b )2 +c 2 + 2a(b ) + 2(b)c + 2ca

= a 2 +b2 + c 2 2ab 2bc + 2ca

Progress Check

Expand the following and verify :

(a + b + c)2 = (abc)2

(a +b +c)2 = (abc)2

(a b +c)2 = (a + bc)2

(a + b c)2 = (ab +c)2

Example 3.18

Expand (2x + 3y + 4z)2

Solution

We know that,

(a + b +c )2 = a 2 +b 2 + c 2 + 2ab + 2bc + 2ca

Substituting, a = 2x , b = 3y andc = 4z

(2x + 3y + 4z)2 = (2x )2 +(3y)2 + (4z )2 + 2(2x )(3y ) + 2(3y)(4z ) + 2(4z )(2x)

= 4x 2 + 9y2 + 16z 2 +12xy + 24yz +16xz

Example 3.19

Find the area of square whose side length is 3m + 2n 4l

Solution

Area of square = side × side

= (3m + 2n4l)×(3m + 2n4l)

= (3m + 2n4l)2

We know that, (a + b +c)2 = a2 + b 2 +c2 + 2ab + 2bc + 2ca

[ 3m + 2n + (4l)]2 = (3m)2 +(2n)2 + (4l)2 + 2(3m)(2n) + 2(2n)(4l ) + 2(4l)(3m)

= 9m2 + 4n2 + 16l2 +12mn16ln24lm

Therefore, Area of square = [9m2 + 4n2 + 16l2 +12mn 16ln 24lm] sq.units.

Substituting

a = 3m,

b = 2n

c = –4l

## 2. Identities involving Product of Three Binomials

( x+ a )(x +b)(x +c) = [(x +a) (x + b )] (x +c)

= [x 2 + (a +b)x +ab](x +c)

= x 2 (x) +(a + b )(x)(x) +abx + x 2c +(a + b )(x)c +abc

= x 3 +ax 2 + bx 2 +abx + cx2 +acx + bcx +abc

= x 3 +(a + b +c)x 2 +(ab + bc +ca)x +abc

Thus, (x + a )(x +b)(x + c ) x 3 + (a +b + c )x 2 +(ab +bc + ca)x +abc

Example 3.20

Expand the following:

(i) (x + 5)(x + 6)(x + 4) (ii ) (3x − 1)(3x + 2)(3x − 4)

Solution

We know that (x + a )(x +b)(x +c) = x 3 +(a + b +c)x 2 + (ab +bc + ca)x +abc --(1)

(i) (x + 5)(x + 6)(x + 4)

Replace: a by 5, b by 6, c by 4 in (1)

= x 3 +(5 + 6 + 4)x2 + (30 + 24 + 20)x +(5)(6)(4)

= x 3 +15x2 + 74x +120

(ii) (3x 1)(3x + 2)(3x 4)

Replace :  x by 3x, a by –1,  b by 2, c by –4 in (1)

= (3x)3 +( 1 + 2 4)(3x)2 +(2 8 + 4)(3x) + (1)(2)(4)

= 27x3 +(− 3)9x2 +(− 6)(3x) + 8

= 27x3 −27x2 − 18x + 8

## 3. Expansion of (x +y)3 and (x -y)3

(x + a )(x +b)(x +c) x 3 +(a + b +c)x2 + (ab +bc + ca )x +abc

substituting a = b = c = y in the identity

we get, (x + y )(x + y )(x + y) = x3 +(y + y +y)x2 + (yy +yy + yy )x + yyy

= x 3 + (3y)x 2 + (3y2 )x +y3

Thus, (x + y)3 x3 + 3x2y + 3xy2 +y3 (or) (x + y)3 x3 +y3 + 3xy(x +y) by replacing y by -y, we get

(x y )3 x33x2y + 3xy2y3 (or) (x y )3 x3y33xy(xy)

Example 3.21

Expand (5a − 3b)3

Solution

We know that,

(x -y)3= x 33x2y + 3xy 2y3

(5a − 3b)3 = (5a )33(5a)2(3b ) + 3(5a)(3b )2 – (3b)3

= 125a 3 3(25a2 )(3b ) + 3(5a)(9b 2 ) – (3b)3

= 125a 3225a2b + 135ab 227b3

The following identity is also used:

x 3 + y 3 + z 3 3xyz (x + y + z )(x2 + y2 + z2 xy yz zx)

We can check this by performing the multiplication on the right hand side.

Note

(i) If (x + y + z) = 0 then x 3 + y 3 + z 3 = 3xyz

Some identities involving sum, difference and product are stated without proof

(i) x3 + y3 (x + y )3 3xy(x + y) (ii) x3 y3 (x y )3 + 3xy(x y)

Example 3.22

Find the product of (2x + 3y + 4z )(4x2 + 9y 2 +16z 2 6xy 12yz 8zx)

Solution

We know that, (a + b +c )(a2 + b 2 +c2 ab bc ca) = a 3 +b3 + c 3 3abc

(2x + 3y + 4z )(4x 2 + 9y 2 +16z 2 6xy 12yz 8zx)

= (2x )3 +(3y)3 + (4z )33(2x )(3y )(4z)

= 8x 3 + 27y 3 + 64z 372xyz

Example 3.23

Evaluate 103 153 + 53

Solution

We know that, if a + b +c = 0 , then a 3 + b 3 +c3 = 3abc

Here, a + b +c =1015+5 = 0

Therefore, 103 + (15)3 + 53 = 3(10)(15)(5)

103 153 +53 = −2250

Replace a by 10, b by –15, c by 5

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