Question

Utiliza los productos notables y resuelve.


(x3+3)2=

(x3-3)2=

(a3-b2)(a3+b2)=

(1-8xy)(1+8xy)=​

Answer 1

Respuesta:

(x3+3)2=

=(x3+3)(x3+3)

=(x3)(x3)+(x3)(3)+(3)(x3)+(3)(3)

=x6+3x3+3x3+9

=x6+6x3+9

(x3-3)2=

=(x3+−3)(x3+−3)

=(x3)(x3)+(x3)(−3)+(−3)(x3)+(−3)(−3)

=x6−3x3−3x3+9

=x6−6x3+9

(a3-b2)(a3+b2)=

=(a3+−b2)(a3+b2)

=(a3)(a3)+(a3)(b2)+(−b2)(a3)+(−b2)(b2)

=a6+a3b2−a3b2−b4

=a6−b4

(1-8xy)(1+8xy)=​

=(1+−8xy)(1+8xy)

=(1)(1)+(1)(8xy)+(−8xy)(1)+(−8xy)(8xy)

=1+8xy−8xy−64x2y2

=−64x2y2+1

Explicación paso a paso:

Answer 2

Respuesta:

a) (x3+3)2=

=(x3+3)(x3+3)

=(x3)(x3)+(x3)(3)+(3)(x3)+(3)(3)

=x6+3x3+3x3+9

=x6+6x3+9

b) (x3-3)2=

=(x3+−3)(x3+−3)

=(x3)(x3)+(x3)(−3)+(−3)(x3)+(−3)(−3)

=x6−3x3−3x3+9

=x6−6x3+9

c) (a3-b2)(a3+b2)=

=(a3+−b2)(a3+b2)

=(a3)(a3)+(a3)(b2)+(−b2)(a3)+(−b2)(b2)

=a6+a3b2−a3b2−b4

=a6−b4

d) (1-8xy)(1+8xy)=​

=(1+−8xy)(1+8xy)

=(1)(1)+(1)(8xy)+(−8xy)(1)+(−8xy)(8xy)

=1+8xy−8xy−64x2y2

=−64x2y2+1

Explicación paso a paso: