Find the Product

Question

(2x+1)³

Answer 1

Are you familiar with Paschal's triangle for the coefficients of a binomial to the nth power?  I will assume you are.  If you are not, then you will have to expand this the long way as

 

(2x + 3)(2x + 3)(2x + 3) = ?

 

Paschal's triangle up to n = 3:

 

1

1 1

1 2 1

1 3 3 1

 

The last line gives you the coefficients.  The powers start with the first term to nth, descending by 1, and the second term to 0th, ascending by 1.

 

(2x + 1)³ = 1(2x)³(1)⁰ + 3(2x)²(1)¹ + 3(2x)¹(1)² + 1(2x)⁰(1)³ = 8x³ + 12x² + 6x + 1

 

 

Answer 2

Will solve this problem 3 different ways:

 

Option #1: brute force, plain, ordinary algebra

 

 

(2x+1)(2x+1)(2x+1) = 

 

(4x^2 + 4x + 1)(2x + 1) =

 

8x^3 + 8x^2 + 2x 

            4x^2 + 4x + 1

 

8x^3 + 12x^2 + 6x + 1

 

 

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Option #2:

you can also use the binomial theorem:

 

 (3 choose 0) * (2x)^3 (1)^0  +

 (3 choose 1) * (2x)^2 (1)^1  +  

(3 choose 2) * (2x)^1 (1)^2   +

(3 choose 3)*(2x)^0*(1)^3

 

(3 choose 0), (3 choose 3), 1 to any power, and any number to the power zero are all 1.

 

(3 choose 1) is 3!/(1! * 2!) = 3*2*1/(1*2*1) = 3

(3 choose 2) is 3!/(2! * 1!) = 3!/(1!*2!) = 3 <--- same calculation

 

The expansion simplifies to:

 

(2x)^3 + 3*(2x)^2 + 3*(2x) + 1

 

8x^3 + 3*(4x^2) + 3*(2x) + 1

 

8x^3 + 12x^2 + 6x + 1

 

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Option #3:

 For a more artistic approach, you can use Pascal's Triangle

 

     1 
  1 2 1
1 3  3 1

(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3

 

But for this problem,  a=2x and  b=1

(2x)^3 + 3(2x)^2(1) + 3(2x)(1) + 1^3

8x^3 + 12x^2 + 6x + 1