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# Application of Cubic Identities

Cubic Identities: Numerical Example solved problems

Application of Cubic Identities

I. Using the identity (a + b)3 = a 3 + 3a2 b + 3ab 2 + b3

Example 3.13

Expand (x + 4)3

Solution:

Comparing (x + 4)3 with (a + b)3 , we get a = x , b = 4

We know (a + b)3 = a 3 + 3a2b + 3ab 2 + b3

(x + 4)3 = (x)3 + 3(x )2 (4) + 3(x)(4)2 + (4)3 (replacing a, b values)

= ( x )3 + 3x2 (4) + 3(x)(16) + 64

(x + 4)3 = x3 + 12x 2 + 48x + 64

(4)2 = 4 × 4 = 16

(4)3 = 4 × 4 ×4=64

Try to expand this by using

(a + b)3 = a3 + b3 + 3ab(a + b)

Example 3.14

Find the value of (103)3

Solution:

Now, (103)3 = (100 + 3)3

Comparing this with ( a + b)3 , we get a = 100 , b = 3

(a + b)3 = a 3 + 3a2b + 3ab2 + b3 replacing a, b values,

(100 + 3)3 = (100)3 + 3(100)2 (3) + 3(100)(3)2 + (3)3

= 1000000+ 3(10000)(3) + 3(100)(9) + 27

= 1000000+ 90000 + 2700 + 27

(103)3 =1092727

II. Using the identity (a - b)3 = a 3 - 3a2 b + 3ab 2 - b3

Example 3.15

Expand: ( y 5)3

Solution:

Comparing ( y 5)3 with (a b)3 , we get a = y , b = 5

(ab)3 = a3 3a2b + 3ab2 b3

(y –5)3 = ( y)3 – 3( y)2 (5) + 3( y)(5)2 - (5)3

= ( y)3 – 3y2 (5) + 3( y)(25) – 125

(y–5)3 = y315y2 + 75y125

Try to expand this by using

(a b)3 = a3 b3 3ab(a b)

Example 3.16

Find the value of (98)3

Solution:

Now, (98)3 = (100 2)3

Comparing this with ( a b)3 , we get a = 100 , b = 2

(a b)3 = a3 3a2b + 3ab2 b3

(100 2)3 = (100)3 3(100)2 (2) + 3(100)(2)2 – (2)3

= 1000000 3(10000)(2) + 3(100)(4) 8

= 1000000 60000 +1200 8

(98)3 = 941192

III. Using the identity (x + a)(x + b )(x + c) = x3 +(a + b + c)x2 +(ab + bc + ca )x + abc

Example 3.17

Expand: (x + 3)(x + 5)(x + 2)

Solution:

Given

(x + 3)(x + 5)(x + 2)

Comparing this with (x + a)(x + b)(x + c) , we get x = x , a = 3 , b = 5 , c = 2

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

(x + 3)(x + 5)(x + 2) = ( x )3 + (3 + 5 + 2)(x)2 + (3 × 5 + 5 × 2 + 2 × 3)x + (3)(5)(2)

= x 3 + 10x2 + (15 + 10 + 6)x + 30

(x + 3)(x + 5)(x + 2) = x3 + 10x2 + 31x + 30

Try these

Expand : (i) ( x + 5)3 (ii) ( y 2)3 (iii) (x + 1)(x + 4)(x + 6)

Solution:

(i) Comparing (x + 5)3 with (a + b)3, we have a = x and b = 4.

(a + b)3 = a3 + 3a2 b + 3ab2 + b3

(x + 5)3 = x3 + 3x2 (5) + 3 (x) (5)2 + 53

. = x3 + 15x2 + 75x + 125

(ii) Comparing (y − 2)3 with (a − b)3 we have a = y b = z

(a − b)3 = a3 − 3a2 b + 3ab2b3

(y − 2)2 = y3 – 3y2(2) + 3y(2)2 + 23

= y3 − 6y2 + 12y + 8

(iii) Comparing (x + 1) (x + 4) (x + 6) with (x + a) (x + b) (x + c) we have

a = 1 b =  4 and c = 6

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

= x3 + (1 + 4 + 6) x2 + ((1) (4) + (4) (6) + (6) (l))x + (1) (4) (6)

= x3 + 11x2 + (4 + 24 + 6) x +24

= x3 + 11x2 + 34x + 24

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