# EXAMPLE PROBLEMS BASED ON CUBES OF BINOMIAL

## About "Example Problems Based on Expansion of Cubes of Binomial"

Example Problems Based on Expansion of Cubes of Binomial :

Here we are going to see some example problems to show how to expand cubes of binomial.

(a + b)3  =  a3 + 3a2b + 3ab2 + b3

(a - b)3  =  a3 - 3a2b + 3ab2 - b3

(a3 + b3)  =  (a + b)3 - 3ab(a + b)

(a3 - b3)  =  (a - b)3 - 3ab(a - b)

## Example Problems Based on Expansion of Cubes of Binomial - Examples

Question 1 :

Expand :

(i) (3a - 4b)3

Solution :

(a - b) =  a3 - 3a2b + 3ab2 - b3

(3a - 4b)3  =  (3a)3 - 3(3a)2(4b) + 3(3a) (4b)2 - (4b)3

=  27a3 - 3(9a2)(4b) + 9a(16b2) - 64b3

=  27a3 - 108a2b + 144ab2 - 64b3

(ii)  (x + (1/y))3

Solution :

(a + b) =  a3 + 3a2b + 3ab2 + b3

(x + (1/y))=  x3 + 3x2(1/y) + 3x(1/y)2 + (1/y)3

=  x3 + (3x2/y) + (3x/y2) + (1/y3)

Question 2 :

Evaluate the following by using identities: (i) 983

Solution :

98 =  (100 - 2)3

(a - b) =  a3 - 3a2b + 3ab2 - b3

=  (100)3 - 3(100)2(2) + 3(100) (2)2 - 23

=  1000000 - 6(10000) + 12(100) - 8

=  1000000 - 60000 + 1200 - 8

=  941192

(ii)  10013

1001=  (1000 + 1)3

(a + b) =  a3 + 3a2b + 3ab2 + b3

=  (1000)3 + 3(1000)2(1) + 3(1000) (1)2 + 13

=  1000000000 + 3(1000000) + 3000 + 1

=  1003003001

Question 3 :

If (x + y + z) = 9 and (xy + yz + zx) = 26, then find the value of x2 + y2 + z2

Solution :

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

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

x2 + y2 + z2  = (9)2 - 2(26)

=  81 - 52

= 29

Hence the value of x2 + y2 + z2 is 29.

Question 4 :

Find 27a3 + 64b3 , if 3a + 4b = 10 and ab = 2

Solution :

27a3 + 64b3   =  33a3 + 43b3  =  (3a)3 + (4b)3

a3 + b =  (a + b)3 - 3ab(a + b)

(3a)3 + (4b) =  (3a + 4b)3 - 3(3a)(4b)(3a + 4b)

=  103 - 36 ab(3a + 4b)

=  1000 - 36(2)(10)

=  1000 - 720

=  280 After having gone through the stuff given above, we hope that the students would have understood, "Example Problems Based on Expansion of Cubes of Binomial"

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