khai triển hăng đẳng thức sau:
(1-x)(1+x)(1+x^2)
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\(a,\left(x+2\right)^2=x^2+4x+4\\ b,\left(x-1\right)^2=x^2-2x+1\\ c,\left(x^2+y^2\right)^2=x^4+2x^2y^2+y^4\)

x 2 - 1 = x 2 - 1 2 = (x + 1) (x - 1)
(x + 1) 3 = x 3 + 3x 2 + 3x + 1
x 3 + 1 = x 3 + 1 3 = (x + 1) (x 2 - x + 1)
(x + 2) 2 = x 2 + 4x + 4
(x - 2) 2 = x 2 - 4x + 4
Hok tốt ~

1) \(x^3-1=x^3-1^3=\left(x-1\right)\left(x^2+x+1\right)\)
2) \(27x^3-64=\left(3x\right)^3-4^3=\left(3x-4\right)\left(9x^2+12x+4\right)\)
3) \(8x^3+1=\left(2x\right)^3+1^3=\left(2x+1\right)\left(4x^2-2x+1\right)\)

\(a,=x^2+4x+4\\ b,=x^3+3x^2+3x+1\\ c,=\left(x-3\right)\left(x+3\right)\)
a,\(\left(x+2\right)^2=x^2+2.x.2+2^2=x^2+4x+4\)
b, \(\left(x+1\right)^3=x^3+3.x^2.1+3.x.1^2+1^3=x^3+3x^2+3x+1\)
c,\(x^2-3^2=\left(x-3\right).\left(x+3\right)\)

Ta khai triển :
\(\left(\right. 1 - x \left.\right) \left(\right. 1 + x \left.\right) = 1 - x^{2}\)
\(\left(\right. 1 + x^{2} \left.\right)\):
\(\left(\right. 1 - x^{2} \left.\right) \left(\right. 1 + x^{2} \left.\right) = 1 - x^{4}\)
Vậy
\(\left(\right. 1 - x \left.\right) \left(\right. 1 + x \left.\right) \left(\right. 1 + x^{2} \left.\right) = 1 - x^{4}\)
\(\left(1-x\right)\left(1+x\right)\left(1+x^2\right)\)
\(=\left(1-x^2\right)\left(1+x^2\right)\)
\(=1^2-\left(x^2\right)^2=1-x^4\)