a plus b WHOLE CUBE FORMULA

In this section, you will learn the formula or expansion  for (a + b)3.

That is, 

(a + b)3  =  (a + b)(a + b)(a + b)

Multiply (a + b) and (a + b).

(a + b)3  =  (a2 + ab + ab + b2)(a + b)

Simplify.

(a + b)3  =  (a2 + 2ab + b2)(a + b)

(a + b)3  =  a3 + a2b + 2a2b + 2ab2 + ab2 + b3

Combine the like terms. 

(a + b)3  =  a3 + 3a2b + 3ab2 + b3

or

(a + b)3  =  a3 + b3 + 3ab(a + b)  

Solved Problems

Problem 1 :

Expand : 

(x + 2)

Solution :

(x + 2)is in the form of (a + b)3

Comparing (a + b)and (x + 2)3, we get

a  =  x

b  =  2

Write the formula / expansion for (a + b)3.

(a + b)3  =  a3 + 3a2b + 3ab2 + b3

Substitute x for a and 2 for b. 

(x + 2)3  =  x3 + 3(x2)(2) + 3(x)(22) + 23

(x + 2)3  =  x3 + 6x2 + 3(x)(4) + 8

(x + 2)3  =  x3 + 6x2 + 12x + 8

So, the expansion of (x + 2)3 is

x3 + 6x2 + 12x + 8

Problem 2 :

Expand : 

(2x + 3)

Solution :

(2x + 3)is in the form of (a + b)3

Comparing (a + b)and (2x + 3)3, we get

a  =  2x

b  =  3

Write the formula / expansion for (a + b)3.

(a + b)3  =  a3 + 3a2b + 3ab2 + b3

Substitute 2x for a and 3 for b. 

(2x + 3)3  =  (2x)3 + 3(2x)2(3) + 3(2x)(32) + 33

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

(2x + 3)3  =  8x3 + 36x2 + 54x + 27

So, the expansion of (2x + 3)3 is

8x3 + 36x2 + 54x + 27

Problem 3 :

Expand : 

(p + 2q)

Solution :

(p + 2q)is in the form of (a + b)3

Comparing (a + b)and (p + 2q)3, we get

a  =  p

b  =  2q

Write the formula / expansion for (a + b)3.

(a + b)3  =  a3 + 3a2b + 3ab2 + b3

Substitute p for a and 2q for b. 

(p + 2q)3  =  p3 + 3(p2)(2q) + 3(p)(2q)2 + (2q)3

(p + 2q)3  =  p3 + 6p2q + 3(p)(4q2) + 8q3

(p + 2q)3  =  p3 + 6p2q + 12pq2 + 8q3

So, the expansion of (p + 2q)3 is

p3 + 6p2q + 12pq2 + 8q3

Problem 4 : 

If a + b  =  12 and a3 + b3  =  468, then find the value of ab. 

Solution :

To find the value of ab, we can use the formula or expansion for (a + b)3.

Write the formula / expansion for (a + b)3.

(a + b)3  =  a3 + 3a2b + 3ab2 + b3

or

(a + b)3  =  a3 + b3 + 3ab(a + b)

Substitute 12 for (a + b) and 468 for (a3 + b3). 

(12)3  =  468 + 3(ab)(12)

Simplify.

1728  =  468 + 36ab

Subtract 468 from each side. 

1260 =  36ab

Divide each side by 36. 

35  =  ab

So, the value of ab is 35. 

Problem 5 :

Find the value of :

(107)3

Solution :

We can use the algebraic formula for (a + b)and find the value of (107)easily.

Write (107)in the form of (a + b)3.

(107)3  =  (100 + 7)3

Write the expansion for (a + b)3.

(a + b)3  =  a3 + b3 + 3ab(a + b)

Substitute 100 for a and 7 for b. 

(100 + 7)3  =  1003 + 73 + 3(100)(7)(100 + 7)

(100 + 7)3  =  1000000 + 343 + 3(100)(7)(107)

(100 + 7)3  =  1000000 + 343 + 224700

(107)3  =  1225043

So, the value of (107)3 is

1,225,043

Apart from the stuff explained above, if you would like to learn about more identities in Algebra, 

Please click here

Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. SAT Math Resources (Videos, Concepts, Worksheets and More)

    Jul 11, 24 05:20 PM

    SAT Math Resources (Videos, Concepts, Worksheets and More)

    Read More

  2. SAT Math Videos (Part - 21)

    Jul 11, 24 05:14 PM

    satmath18.png
    SAT Math Videos (Part - 21)

    Read More

  3. Best Way to Learn Mathematics

    Jul 11, 24 11:59 AM

    Best Way to Learn Mathematics

    Read More