\(a.x^3-\left(2y\right)^3\\ =\left(x-2y\right)\left[x^2+x\cdot2y+\left(2y\right)^2\right]\\ =\left(x-2y\right)\left(x^2+2xy+4y^2\right)\\ b.x^3-64\\ =x^3-4^3\\ =\left(x-4\right)\left(x^2+4\cdot x+4^2\right)\\ =\left(x-4\right)\left(x^2+4x+16\right)\\ c.8x^3-y^3\\ =\left(2x\right)^3-y^3\\ =\left(2x-y\right)\left[\left(2x\right)^2+2x\cdot y+y^2\right]\\ =\left(2x-y\right)\left(4x^2+2xy+y^2\right)\\ d.\dfrac{1}{27}x^6-125\\ =\left(\dfrac{1}{3}x^2\right)^3-5^3\\=\left(\dfrac{1}{3}x^2-5\right)\left[\left(\dfrac{1}{3}x^2\right)^2+\dfrac{1}{3}x^2\cdot5+5^2\right]\\ =\left(\dfrac{1}{3}x^2-5\right)\left(\dfrac{1}{9}x^4+\dfrac{5}{3}x^2+25\right)\)
a) $x^3-(2y)^3$
$=(x-2y)[x^2+x.2y+(2y)^2]$
$=(x-2y)(x^2+2xy+4y^2)$
b) $x^3-64$
$=x^3-4^3$
$=(x-4)(x^2+x.4+4^2)$
$=(x-4)(x^2+4x+16)$
c) $8x^3-y^3$
$=(2x)^3-y^3$
$=(2x-y)[(2x)^2+2x.y+y^2]$
$=(2x-y)(4x^2+2xy+y^2)$
d) $\frac{1}{27}x^6-125$
$=\left(\frac13 x^2\right)^3-5^3$
$=\left(\frac13 x^2-5\right)\left[\left(\frac13 x^2\right)^2+\frac13 x^2.5+5^2\right]$
$=\left(\frac13 x^2-5\right)\left(\frac 19x^4+\frac53 x^2+25\right)$
#$\mathtt{Toru}$
