ALGEBRAIC IDENTITIES

An algebraic identity is an equality that remains true regardless of the values chosen for its variables. Algebraic identities can be used to get the expansion of a polynomial for the given exponent.

Expansion of Binomial

(a + b)2 = a2 + 2ab + b2

(a - b)2 = a2 - 2ab + b2

(a + b)(a - b) = a2 - b2

(x + a)(x + b) = x2 + (a + b)x + ab

Expand the following using the algebraic identities above :

Example 1 :

(3x + 4y)2

Solution :

We have

(a + b)2 = a2 + 2ab + b2

Substitute a = 3x and b = 4y.

(3x + 4y)2(3x)2 + 2(3x)(4y) + (4y)2

32x2 + 2(3x)(4y) + 42y2

= 9x2 + 24xy + 16y2

Example 2 :

(2p - 3q)2

Solution :

We have

(a - b)2 = a2 - 2ab + b2

Substitute a = 2p and b = 3q.

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

= 22p2 - 12pq + 32q2

= 4p2 - 12pq + 9q2

Example 3 :

(5x + 4y)(5x - 4y)

Solution :

We have

(a + b)(a - b) = a2 - b2

Substitute a = 5x and b = 4y.

(5x + 4y)(5x - 4y) = (5x)2 - (4y)2

52x2 - 42y2

= 25x2 - 16y2

Example 4 :

(m + 3)(m + 5)

Solution :

We have

(x + a)(x + b) = x2 + (a + b)x + ab

Substitute x = m, a = 3 and b = 5.

(m + 3)(m + 5) = m2 + (3 + 5)m + (3)(5)

= x2 + 8m + 15

Example 5 :

(m - 7)(m + 2)

Solution :

We have

(x + a)(x + b) = x2 + (a + b)x + ab

Substitute x = m, a = -7 and b = 7.

(m - 7)(m + 2) = m2 + (-7 + 2)m + (-7)(2)

= m2 - 5m - 14

Expansion of Trinomial

Expansion of (a + b + c)2 :

We know that

(x + y)2 = x2 + 2xy + y2

Substitute x = a + b and y = c.

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

= a2 + 2ab + b2 + 2ac + 2bc + c2

= a2 + b2 + c+ 2ab + 2bc + 2ac

Thus,

(a + b + c)= a2 + b2 + c+ 2ab + 2bc + 2ac

Example 6 :

Expand (a - b + c)2.

Solution :

We have

(a + b + c)2 = = a2 + b2 + c+ 2ab + 2bc + 2ac

Replace 'b' by '-b'.

(a + (-b) + c)2 = = a2 + (-b)2 + c+ 2a(-b) + 2(-b)c + 2ac

(a - b + c)2 = = a2 + b2 + c- 2ab - 2bc + 2ac

Example 7 :

Expand (a + b - c)2.

Solution :

We have

(a + b + c)2 = = a2 + b2 + c+ 2ab + 2bc + 2ac

Replace 'c' by '-c'.

(a + b + (-c))2 = = a2 + b2 + (-c)+ 2ab + 2b(-c) + 2a(-c)

(a + b + (-c))2 = = a2 + b2 + c+ 2ab - 2bc - 2ac

Example 8 :

Expand (a - b - c)2.

Solution :

We have

(a + b + c)2 = = a2 + b2 + c+ 2ab + 2bc + 2ac

Replace 'b' by '-b' and 'c' by '-c'.

(a + (-b) + (-c))2 = = a2 + (-b)2 + (-c)+ 2a(-b) + 2(-b)(-c) + 2a(-c)

(a - b - c)2 = = a2 + b2 + c- 2ab + 2bc - 2ac

Example 9 :

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

Solution :

We have

(a + b + c)2 = = a2 + b2 + c+ 2ab + 2bc + 2ac

Substitute a = 2x, b = 3y and c = 4z.

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

= 22x2 + 32y2 + 42z+ 12xy + 24yz + 16xz

= 4x2 + 9y2 + 16z+ 12xy + 24yz + 16xz

Example 10 :

Find the area of square whose side length is (3p + 2q - 4r).

Solution :

Area of Square :

= side x side

= (3p + 2q - 4r)(3p + 2q - 4r)

= (3p + 2q - 4r)2

We know that

(a + b + c)2 = = a2 + b2 + c+ 2ab + 2bc + 2ac

Substitute a = 3p, b = 2q and c = -4r.

(3p + 2q + (-4r)2 = (3p)2 + (2q)2 + (-4r)+ 2(3p)(2q) + 2(2q)(-4r) + 2(3p)(-4r)

= 9p2 + 4q2 + 16r+ 12pq - 16qr - 24pr

Therefore, Area of Square = [9p2 + 4q2 + 16r+ 12pq - 16qr - 24pr] square units.

Identities involving Product of Three Binomials

Expansion of (x + a)(x + b)(x + c) :

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

= [x2 + (a + b)x + ab](x + c)

= x2(x) + (a + b)(x)(x) + ab(x) + x2(c) + (a + b)(x)(c) + ab(c)

= x3 + ax2 + bx+ abx + cx2 + acx + bcx + abc

= x3 + (a + b + c)x2 + (ab + bc + ac)x + abc

Thus,

(x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ac)x + abc

Example 11 :

Expand (x + 4)(x + 5)(x + 6).

Solution :

We know that

(x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ac)x + abc

Substitute a = 4, b = 5 and c = 6.

(x + 4)(x + 5)(x + 6) = x3 + (4 + 5 + 6)x2 + [(4)(5) + (5)(6) + (4)(6)]x + (4)(5)(6)

= x3 + 15x2 + (20 + 30 + 24)x + 120

= x3 + 15x2 + 74x + 120

Example 12 :

Expand (3x - 1)(3x + 2)(3x - 4).

Solution :

We know that

(x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ac)x + abc

Substitute x = 3x, a = -1, b = 2 and c = -4.

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

= (3x)3 + (-1 + 2 - 4)(3x)2 + [(-1)(2) + (2)(-4) + (-1)(-4)](3x) + (-1)(2)(-4)

= 27x3 + (-3)(9x2) + (-2 - 8 + 4)(3x) + 8

= 27x3 - 27x2 + (-6)(3x) + 8

= 27x3 - 27x2 - 18x + 8

Cube of a Binomial

Expansion of (x + y)3 :

(x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ac)x + abc

Substituting a = b = c = y.

(x + y)(x + y)(x + y) = x3 + (y + y + y)x2 + (yy + yy + yy)x + yyy

(x + y)3 = x3 + (3y)x2 + (y2y2 + y2)x + y3

= x3 + 3x2y + (3y2)x + y3

= x3 + 3x2y + 3xy2 + y3

or

= x3 + 3xy(x + y)+ y3

Thus,

(x + y)3 = x3 + 3x2y + 3xy2 + y3

or

(x + y)3 = x3 + 3xy(x + y)+ y3

Example 13 :

Expand (x - y)3.

Solution :

We know that

(x + y)3 = x3 + 3x2y + 3xy2 + y3

Replace 'y' by '-y'.

(x + (-y))3 = x3 + 3x2(-y) + 3x(-y)2 + (-y)3

= x3 - 3x2y + 3xy2 - y3

or

=  = x3 - 3xy(x - y) - y3

Example 14 :

Expand (5a - 3b)3.

Solution :

We know that

(x + y)3 = x3 + 3x2y + 3xy2 + y3

Substitute x = 5a, and y = -3b.

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

= 53a3 + 3(25a2)(-3b) + 3(5a)(9b2) + (-27b3)

= 125a3 - 225a2b + 135ab2 - 27b3

Other Identities

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

2) If (a + b + c) = 0, then a3 + b3 + c3 = 3abc

Some identities involving sum, difference and product are stated here :

x3 + y= (x + y)3 - 3xy(x + y)2

x3 - y= (x - y)3 + 3xy(x - y)2

Example 15 :

Find the product of :

(2x + 3y + 4z)(4x2 + 9y2 + 16z2 - 6xy - 12yz - 8xz)

Solution :

We know that

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

Substitute a = 2x, b = 3y and c = 4z.

(2x + 3y + 4z)(4x2 + 9y2 + 16z2  - 6xy - 12yz - 8xz)

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

= 8x3 + 27y3 + 64z3 - 72xyz

Example 16 :

Evaluate : 103 - 153 + 53.

Solution :

We know that, If (a + b + c) = 0, then a3 + b3 + c3 = 3abc.

If a = 10, b = -15 and c = 5, then

a + b + c = 10 - 15 + 5 = 0

Therefore,

103 + (-15)3 + 53 = 3(10)(-15)(5)

103 - 153 + 5= -2250

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