only 2 i cant get
$\displaystyle 4(m+n)^3 - 4 =$
$\displaystyle 4[(m+n)^3 - 1] =$
note the difference of cubes in the brackets ...
$\displaystyle 4[(m+n)^3 - 1^3]$
remember how to factor the difference of two cubes?
$\displaystyle 36x^2-9y^2+6y-1=36x^2 - (9y^2 - 6y + 1)$
$\displaystyle 36x^2 - (3y - 1)^2$
difference of two squares, correct?
Hello, redhawks!
$\displaystyle 4(m+n)^3 - 4$
We have: .$\displaystyle 4\underbrace{\bigg[(m+n)^3 - 1\bigg] }_{\text{diff. of cubes}}$
. . . $\displaystyle = \;\bigg[(m+n)-1\bigg]\,\bigg[(m+n)^2 + (m+n) + 1\bigg]$
. . . $\displaystyle = \;(m+n-1)(m^2 + 2mn + n^2 + m + n + 1)$
$\displaystyle 36x^2 - 9y^2 + 6y - 1$
We have: .$\displaystyle 36x^2 - (9y^2 - 6y + 1)$
. . . $\displaystyle = \;36x^2 - (9y^2 - 6y + 1)$
. . . $\displaystyle = \;\underbrace{36x^2 - (3y - 1)^2}_{\text{diff. of squares}}$
. . . $\displaystyle = \;\bigg[6x - (3y-1)\bigg]\,\bigg[6x + (3y - 1)\bigg]$
. . $\displaystyle = \;(6x - 3y + 1)(6x + 3y - 1)$
Edit: Too slow . . . again!
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