Bài 1 :
$(2x)^2.(4x-2)-(x^3-8x^2)=15$
$⇔4x^2.(4x-2)-(x^3-8x^2)=15$
$⇔16x^3-8x^2-x^3+8x^2=15$
$⇔15x^3=15$
$⇔x^3=1$
$⇔x=1$
$Vậy\ x=1$
Bài 2 :
$a.(x^3-2x^2+x-1)(5-x) \\=5x^3-10x^2+5x-5-x^4+2x^3-x^2+x \\=7x^3-11x^2+6x-5-x^4$
$b.(c+3)(c-2)(c+1) \\=(c^2+3c-2c-6)(c+1) \\=(c^2+c-6)(c+1) \\=c^3+c^2-6c +c^2+c-6\\=c^3+2c^2-5c-6$