Question

(a+b) (a2+ab+b2)

(2a2b+6b) (2b+4a2+6)

multiplicacion

le doy corona el que lo haga promesa ​

Answer 1

Respuesta:

La respuesta es:

$22a^4b^2+16a^3b^3+10a^2b^4+2ab^5+15a^3b^2+18a^2b^3+12ab^4+3b^5+4a^6b+8a^5b^2+4a^3b^4+6a^5b+6a^4b+9a^3b+18a^2b^2+18ab^3+9b^4$

Explicación paso a paso:

Aplicamos la propiedad distributiva (Esto va ha ser Super Largo)

$(a+b)*(a^2+ab+b^2)*(2a*2b+6b)*(2b+4a^2+6)\\ \\ (a^3+a^2b+ab^2+a^2b+ab^2+b^3)*(4ab+6b)*(2b+4a^2+6)\\ \\ (a^3+2a^2b+2ab^2+b^3)*(4ab+6b)*(2b+4a^2+6)\\ \\ (4a^4b+8a^3b^2+8a^2b^3+4ab^4+6a^3b+12a^2b^2+12ab^3+6b^4)*(2b+4a^2+6)\\ \\ \\ (8a^4b^2+16a^3b^3+16a^2b^4+8ab^5+12a^3b^2+24a^2b^3+24ab^4+12b^5+16a^6b+32a^5b^2+32a^4b^2+16a^3b^4+24a^5b+48a^4b^2+48a^3b^3+24a^2b^4+24a^4b+48a^3b^2+48a^2b^3+24ab^4+36a^3b+72a^2b^2+72ab^3+36b^4)$

Sumamos todos los terminos comunes

$88a^4b^2+64a^3b^3+40a^2b^4+8ab^5+60a^3b^2+72a^2b^3+48ab^4+12b^5+16a^6b+32a^5b^2+16a^3b^4+24a^5b+24a^4b+36a^3b+72a^2b^2+72ab^3+36b^4$

Luego dividimos todo entre 4:

$22a^4b^2+16a^3b^3+10a^2b^4+2ab^5+15a^3b^2+18a^2b^3+12ab^4+3b^5+4a^6b+8a^5b^2+4a^3b^4+6a^5b+6a^4b+9a^3b+18a^2b^2+18ab^3+9b^4$