Szczegółowe wyjaśnienie:
Mamy w zadaniach zastosowanie wzorów skróconego mnożenia.
I tak:
Zad.1
[tex]a)\ (a+b)^3=a^3+3a^2b+3ab^2+b^3\\\\b)\ (a-b)^3=a^3-3a^2b+3ab^2-b^3\\\\c)\ (a+b)^3=a^3+3a^2b+3ab^2+b^3[/tex]
Zad.2
[tex]a^3-b^3=(a-b)(a^2+ab+b^2)\\\\a^3+b^3=(a+b)(a^2-ab+b^2)\\\\a^2-b^2=(a-b)(a+b)[/tex]
[tex]Zad.1\\\\a)\ (3x+1)^3=(3x)^3+3\cdot(3x)^2\cdot1+3\cdot3x\cdot1^2+1^3=27x^3+27x^2+9x+1\\\\b)\ \left(2x-\dfrac{1}{2}y\right)^3=(2x)^3-3\cdot(2x)^2\cdot\dfrac{1}{2}y+3\cdot2x\cdot\left(\dfrac{1}{2}y\right)^2-\left(\dfrac{1}{2}y\right)^3\\\\=8x^3-6x^2y+\dfrac{3}{2}xy^2-\dfrac{1}{8}y^3\\\\c)\ (-\sqrt2-2x)^3=\bigg[-(\sqrt2+2x)\bigg]^3=(-1)^3(\sqrt2+2x)^3=-(\sqrt2+2x)^3\\\\=-\bigg[(\sqrt2)^3+3\cdot(\sqrt2)^2\cdot2x+3\cdot\sqrt2\cdot(2x)^2+(2x)^3\bigg]\\\\=-2\sqrt2-12x-12x^2\sqrt2-8x^3[/tex]
[tex]Zad.2\\\\\left(\dfrac{1}{2}x-2y\right)\left(\dfrac{1}{4}x^2+xy+4y^2\right)\left(\dfrac{1}{4}x^2-xy+4y^2\right)\left(\dfrac{1}{2}x+2y\right)\\\\=\left(\dfrac{1}{2}x-2y\right)\bigg[\left(\dfrac{1}{2}x\right)^2+xy+(2y)^2\bigg]\bigg[\left(\dfrac{1}{2}x\right)^2-xy+(2y)^2\bigg]\left(\dfrac{1}{2}x+2y\right)\\\\=\bigg[\left(\dfrac{1}{2}x\right)^3-(2y)^3\bigg]\bigg[\left(\dfrac{1}{2}x\right)^3+(2y)^3\bigg]=\bigg[\left(\dfrac{1}{2}x\right)^3\bigg]^2-\bigg[(2y)^3\bigg]^2[/tex]
[tex]=\left(\dfrac{1}{2}x\right)^6-(2y)^6=\dfrac{1}{64}x^6-64y^6[/tex]