Can't figure out this Math problem?

Depends on whether we should take the 1st equation as written or not.

Mathsteacher to the (x-3) as one quantity, they are not.

So did you want it to be (x-3)/(x^4-81)?

If so, then use the difference of 4th powers formula for the bottom, noting that 81=3^4.

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

So it is (x-3)/((x-3)(x^3+3x^2+9x+27))

=

The (x-3) factors cancel out, leaving

1/(x^3+3x^2+9x+27)

or 1/((x^2+9)(x+3))

Or just factor x^4-81 using the difference of squares formula.

That is a^2-b^2=(a+b)(a-b)

(x^2)^2-9^2=(x^2+9)(x^2-9)

The second factor is itself a difference of squares.

Applying the difference of squares formula, we have (x^2+9)(x-3)(x+3)

This is divided in the original into (x-3). As above, the (x-3) factors cancel out, and we have 1/((x^2+9)(x+3))

x-3/ x^4+0x^3+0x^2+0x-81

= (x-3)/(x^4-81)

= (x-3)/(x^2-9)(x^2+9)

= (x-3)/(x-3)(x+3)(x^2+9)

= 1/(x+3)(x^2+9)

1/(x^3 + 3x^2 +9x +27)

(23x^7n+11)(-3x^5n-20)

= -69x^7nx^5n -460x^7n -33x^5n -220

= -69x^12n - 460x^7n-33x^5n -220