Odpowiedź:
zad.1
(3x-4)²-2(2x+5)²-(x-8)(x+8)=9x²-24x+16-8x²-20x-50-x²+64=30-44x
zad.2
a). (2+b)³=8+12b+6b²+b³
b). (x-2y)³=x³-6x²y+12xy²-8y³
zad.3
a) x³-27=(x-3)(x²+3x+9)
b). 16-x⁴=(4-x²)(4+x²)=(2-x)(2+x)(4+x²)
c). x⁵/32 -1= x⁵/2⁵ -1= (x/2)⁵-1=(x/2 -1)[(x/2)⁴+(x/2)³+(x/2)²+x/2+1]
