Question

desarrolla como producto o cociente notable
a). (-3xy^2-1)^2
b). (ab^3c^2+5ab)^2
c). (5m^2n^3+2)^3
d). x^5-y^5/x-y
e). 8-27m^6/2-3m^2
f). 125y^12+1/5y^4+1

Answer 1

$(a \pm b)^2= a^2\pm 2ab+b^2 \\ \\ a) (-3xy^2-1)^2= (-3xy^2)^2+2(-3xy^2)(-1)+(-1)^2\\ \\ (-3xy^2-1)^2= (-3xy^2)^2+2(3xy^2)+(-1)^2\\ \\ \boxed{(-3xy^2-1)^2= 9x^2y^4+6xy^2+1} \\ \\ \\ b) (ab^3c^2+5ab)^2=(ab^3c^2)^2+2*(ab^3c^2)*(5ab)+(5ab)^2 \\ \\ \boxed{(ab^3c^2+5ab)^2=a^2b^6c^4+10a^2b^4c^2+25a^2b^2}$

$c) (5m^2n^3+2)^3= (5m^2n^3)^3+3(5m^2n^3)^2(2)+3(5m^2n^3)(2)^2+(2)^3 \\ \\ (5m^2n^3+2)^3= (125m^6n^9)+3(25m^4n^6)(2)+3(5m^2n^3)(4)+(8) \\ \\ (5m^2n^3+2)^3= 125m^6n^9+6(25m^4n^6)+12(5m^2n^3)+8 \\ \\ \boxed{(5m^2n^3+2)^3= 125m^6n^9+150m^4n^6+60m^2n^3+8} \\ \\ \\ d) \frac{x^5-y^5}{x-y}= \frac{(x-y)*(y^4+x*y^3+x^2*y^2+x^3*y+x^4)}{x-y}\ simplificamos \\ \\ \boxed{ \frac{x^5-y^5}{x-y}=(y^4+xy^3+x^2y^2+x^3y+x^4)}$


$e) \frac{8-27m^6}{2-3m^2} = \frac{(2-3m^2)(9*m^4+6*m^2+4)}{(2-3m^2)} \ simplificamos \\ \\ \boxed{\frac{8-27m^6}{2-3m^2} = (9*m^4+6*m^2+4)} \\ \\ \\ f) \frac{125y^{12}+1 }{5y^4+1}= \frac{(5*y^4+1)*(25*y^8-5*y^4+1)}{5y^4+1} \\ \\ \frac{125y^{12}+1 }{5y^4+1}= \frac{(1)*(25*y^8-5*y^4+1)}{1} \ simplificamos \\ \\ \boxed{\frac{125y^{12}+1 }{5y^4+1}= 25y^8-5y^4+1}$

Espero que te sirva, salu2!!!!