Question

(a−9) 3 −64
разложите на множители
3(это в кубе)

Answer 1

$\boxed {\; x^3-y^3=(x-y)(x^2+xy+y^2)\; }\\\\\\(a-9)^3-64=(a-9)^3-4^3=\\\\=\Big((a-9)-4\Big)\Big((a-9)^2+4(a-9)+16\Big)=\\\\=(a-13)(a^2-18a+81+4a-36+16)=\\\\=(a-13)(a^2-14a+61)$