Question

How do you expand `(x^2+3)^6`?

Answer 1

`(x^2+3)^6 =x^12 + 18 x^10 + 135 x^8 + 540 x^6 + 1215 x^4 + 1458 x^2 + 729`

Explanation:

It's easy to generate the `1\ 6 ...` line of Pascal's Triangle, which is the one we need. We have ones along the edges and each number is the sum of the two above it.

              1
            1  1
         1   2  1
      1   3   3  1
   1   4    6   4 1
  1  5  10 10  5 1 
 1 6 15 20 15 6 1

Now we can just write

`(a+b)^6 = a^6 + 6 a^5 b + 15 a^4 b^2 + 20 a^3 b^3 + 15a^2b ^4 + 6ab^5 + b^6`

Note how the exponents on `a` count down and the ones on `b` count up so every term is sixth degree.

In our problem we have `a=x^2` and `b=3` so we substitute:

`(x^2+3)^6 = (x^2)^6 + 6 (x^2)^5 (3) + 15 (x^2)^4 3^2 + 20 (x^2)^3 3^3 + 15(x^2)^2 3^4 + 6(x^2)3^5 + 3^6`

Some arithmetic later,

`(x^2+3)^6 =x^12 + 18 x^10 + 135 x^8 + 540 x^6 + 1215 x^4 + 1458 x^2 + 729`

Answer 2

The answer is `=x^12+18x^10+135x^8+540x^6+1215x^4+1458x^2+729`

Explanation:

Apply the binomial expansion (or you can use Pascal's triangle)

`(a+b)^n=sum_(k=0)^n((n),(k))a^kb^(n-k)`

Where,

`((n),(k))=(n!)/((k!)(n-k)!)`

Here,

`a=x^2`

`b=3`

and

`n=6`

`((6),(0))=1`

Therefore,

`(x^2+3)^6=(x^2)^6+6(x^2)^5 3+15(x^2)^4 3^2 + 20(x^2)^3 3^3+15(x^2)^2 3^4+ 6(x^2)3^5+3^6`

`=x^12+18x^10+135x^8+540x^6+1215x^4+1458x^2+729`