Question

Can someone help me with #2? Please explain what you did.

Answer 1

$\bf 64x^6-27y^3\qquad \begin{cases} 64x^2\implies 2^6x^6\implies (2^2x^2)^3\implies (4x^2)^3\\ ----------\\ 27y^3\implies 3^3y^3\implies (3y)^3 \end{cases} \\\\\\ (4x^2)^3-(3y)^3\\\\ -----------------------------\\\\ \textit{difference of cubes} \\ \quad \\ a^3+b^3 = (a+b)(a^2-ab+b^2)\qquad (a+b)(a^2-ab+b^2)= a^3+b^3 \\ \quad \\ a^3-b^3 = (a-b)(a^2+ab+b^2)\qquad (a-b)(a^2+ab+b^2)= a^3-b^3\\\\ -----------------------------$

$\bf (4x^2-3y)[(4x^2)^2+(4x^2)(3y)+(3y)^2] \\\\\\ (4x^2-3y)(16x^4+12x^2y+9y^2)$