Question

6b^3+48b/b^3+64-3b^2/b^2-4b+16

Answer 1

$boldsymbol{dfrac{6b^3+48b}{b^3+64}-dfrac{3b^2}{b^2-4b+16}=}\\\=dfrac{6b^3+48b}{(b+4)(b^2-4b+16)}-dfrac{~~~3b^2^{~~~~~(underline{b+4}}}{b^2-4b+16}=\\\=dfrac{6b^3+48b}{(b+4)(b^2-4b+16)}-dfrac{3b^3+12b^2}{(b+4)(b^2-4b+16)}=\\\=dfrac{6b^3+48b-3b^3-12b^2}{(b+4)(b^2-4b+16)}=dfrac{3b^3-12b^2+48b}{(b+4)(b^2-4b+16)}=\\\=dfrac{3b(b^2-4b+16)}{(b+4)(b^2-4b+16)}boldsymbol{=dfrac{3b}{b+4}}$

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Использована формула суммы кубов

a³ + b³ = (a + b)(a² - ab + b²)