(3x+4y)(3x-4y)

** Anytime there is no middle term you know that the values inside the parenthesis must be + and –

j^3-27=

(j-3)^3

** Since all the terms in the equation have cube roots you can take out the power of 3 and leave it outside the parenthesis.

m^3-125 =

(m-5)^3

** Same as the one above

9x^2-16y^2 = (3x + 4y)(3x – 4y)

2. Another standard formula: a^3 – b^3 = (a – b)(a^2 + ab + b^2) so

j^3 – 27 = (j – 3)(j^2 + 3j + 9)

3. works like 2.

That solves all three of your “2” problems. Reminds me of the old joke, “There are 3 kinds of mathematicians: those who can count and those who can’t.” 🙂

This can be re-written as (3x)^2-(4y)^2 . See why? Becasue 9=3^2, 16=4^2. Then we apply the difference of squares formula, (a^2-b^2)=(a-b)(a+b). Thus in factored form we have (3x-4y)(3x+4y). Multiply it out, it works right?

2. rewrite 27 as 3^3. Then apply difference of cubes formula

(a – b)(a^2 + ab + b^2), yielding (j-3)(j^2+3j+3^2).

3. same process as #2

BTW Willy’s answer is wrong. a^3-b^3 certainly does not equal (a-b)^3