Usiamo le identità A3+B3=(A+B)(A2−AB+B2) e A3−B3=(A−B)(A2+AB+B2).
(a) b3+64=b3+43=(b+4)(b2−4b+16).
(b) 8x3+27y3=(2x)3+(3y)3=(2x+3y)(4x2−6xy+9y2).
(c) Raccogliamo prima 2: 2r3−54=2(r3−27)=2(r−3)(r2+3r+9).
(d) 1000−x6y3=103−(x2y)3=(10−x2y)(100+10x2y+x4y2).
(b+4)(b2−4b+16) ; (2x+3y)(4x2−6xy+9y2) ; 2(r−3)(r2+3r+9) ; (10−x2y)(100+10x2y+x4y2)