\(A=27x^3+1-\left(1-27x^3\right)\\ A=27x^3+1-1+27x^3=54x^3\\ A=54\cdot10^3=54000\)

A = (3x + 1)(9x² - 3x + 1) - (1 - 3x)(1 + 3x + 9x²) 

A = (27x³ + 1) - (1 - 27x³)

A = 27x³ + 1 - 1 + 27x³

A = 54x³

Thay x = 10 vào, ta được:

A = 54 . 10³

A = 54 . 1000

A = 54000

A = (3x + 1)[(3x)² - 3x + 1²] - (1 - 3x)[1² + 3x + (3x)²]

= (3x)³ + 1³ - [1³ - (3x)³]

= 27x³ + 1 - 1 + 27x³ = 54x³

Tại x = 10, giá trị của A là:

54 . 10³ = 54000

\(a,=\left(x+3\right)^3=\left(-3+3\right)^3=0\\ b,=27x^3+1-\left(1-27x^3\right)=27x^3+1-1+27x^3=54x^3\\ =54\cdot10^3=54\cdot1000=54000\)

c, hình như sai đề á e

\(A=\left(X+3\right)^3\)

Thay x = - 3 vào, ta có:

\(A=\left(-3+3\right)^3=0\)

\(a,=\left(3x-11\right)\left(3x+11\right)\\ b,=\left(3x+1-x+2\right)\left(3x+1+x-2\right)\\ =\left(2x+3\right)\left(4x-1\right)\\ c,=\left(2x+1-8\right)\left(2x+1+8\right)=\left(2x-7\right)\left(2x+9\right)\)

\(d,\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=15\\ \Leftrightarrow24x=-10\Leftrightarrow x=-\dfrac{5}{12}\\ e,\Leftrightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x=17\\ \Leftrightarrow9x=10\Leftrightarrow x=\dfrac{10}{9}\\ f,\Leftrightarrow9x^2+18x+9-18x=36+x^3-27\\ \Leftrightarrow x^3-9x^2=0\Leftrightarrow x^2\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\)

Lời giải:

 Áp dụng HĐT: $(a-b)^3=a^3-b^3-3ab(a-b)$ cho cả hai bạn.

a. 

$M=x^3-1-3x(x-1)-3x(x-1)^2+3x^2(x-1)+x^3$
$=2x^3-1+3x(x-1)[-1-(x-1)+x]$

$=2x^3-1+3x(x-1).0=2x^3-1$

b.

$D=[(x-y)-x]^3=-y^3$

c: Ta có: \(x^3-12x^2+48x-64=0\)

\(\Leftrightarrow x-4=0\)

hay x=4

c: Ta có: \(x^3-12x^2+48x-64=0\)

\(\Leftrightarrow x-4=0\)

hay x=4