Question

Question #ea843

Answer 1

`729x^12-1458x^9+1215x^6-540x^3-18/x^3+1/x^6+135`

Explanation:

To expand powers of a sum (or a difference), you can use Pascal's triangle:

The triangle must be read in this way: the first line represents the coefficient of the expansion of `(a+b)^0`, which equals `1`.

The second line represents the coefficient of the expansion of `(a+b)^1`, which equals `a+b`. So, as you can see, the coefficients are `1`

The thirdrepresents the coefficient of the expansion of `(a+b)^2`, which equals `a^2+2ab+b^2`. The coefficients are in fact `1`, `2` and `1`, and the powers of `a` and `b` appear with this pattern: the exponents always sum up to `2`, and at the beginning its `a^2b^0`, then there is `a^1b^1`, and finally `a^0b^2`. So, the exponents of `a` are decreasing, and the ones of `b` are increasing.

Now let's examine `(a+b)^6`: the coefficients will be given by the seventh line, which is

`1\ \ 6\ \ 15\ \ 20\ \ 15\ \ 6\ \ 1`

And the powers will be

`a^6\ \ a^5b\ \ a^4b^2\ \ a^3b^3\ \ a^2b^4\ \ ab^5\ \ b^6`

This means that we have that `(a+b)^6` is

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

In your case, `a=3x^2` and `b=-1/x`. So, you have that:

• `a^6 = 729x^12`;
• `6a^5b =1458x^10*(-1/x) = -1458x^9`;
• `15a^4b^2 = 1215x^8*1/x^2 = 1215x^6`;
• `20a^3b^3 = 540x^6 * (-1/x^3) = -540x^3`;
• `15a^2b^4 = 135x^4 * 1/x^4 = 135`;
• `6ab^5 = 18x^2 * (-1/x^5) = -18/x^3`;
• `b^6 = 1/x^6`

Sum them up and you've got the solution.