Question

Помогите пожалуйста срочно​

Answer 1

Ответ:

$\bigg (\dfrac{1}{2}-x \bigg ) \bigg (\dfrac{1}{4}+\dfrac{1}{2}x+x^{2} \bigg );$

$\bigg (\dfrac{2}{3}-x^{4} \bigg ) \bigg (\dfrac{4}{9}+\dfrac{2}{3}x^{4}+x^{8} \bigg );$

Объяснение:

$a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2});$

$\bigg (\dfrac{a}{b} \bigg )^{n}=\dfrac{a^{n}}{b^{n}}; \quad (a^{b})^{c}=a^{b \cdot c};$

$-x^{6}+\dfrac{1}{8}=\dfrac{1^{3}}{2^{3}}-x^{2 \cdot 3}=\bigg (\dfrac{1}{2} \bigg )^{3}-(x^{2})^{3}=\bigg (\dfrac{1}{2}-x \bigg ) \bigg (\dfrac{1}{4}+\dfrac{1}{2}x+x^{2} \bigg );$

$-x^{12}+\dfrac{8}{27}=\dfrac{2^{3}}{3^{3}}-x^{4 \cdot 3}=\bigg (\dfrac{2}{3} \bigg )^{3}-(x^{4})^{3}=\bigg (\dfrac{2}{3}-x^{4} \bigg ) \bigg (\dfrac{4}{9}+\dfrac{2}{3}x^{4}+x^{8} \bigg );$