Question

desarrollamos el cubo de un binomio a). (10+x)³= b). (y-4)³= c). (mn+1 )³= d). (2-xy)³=​

Answer 1

Respuesta:

a). (10+x)³=  

$10^{3} + 3(10) ^{2} (x) + 3(10)(x )^{2} + x^{3} \\\\1000 +300x + 30x^{2} + x^{3}\\\\ordenas\\\\x^{3} + 30x^{2} +300x +1000$

b). (y-4)³=

$y^{3} - 3(y)^{2} (-4) + 3(y)(4)^{2} - (-4)^{3} \\\\y^{3} + 12y^{2} + 48y + 64$

c). (mn+1 )³=

$(mn)^{3} + 3(mn) ^{2} (1) + 3(mn)(1)^{2} + 1^{3}\\\\m^{3} m^{3} + 3m^{2} n^{2} + 3mn + 1$

d). (2-xy)³=​

$2^{3} - 3(2)^{2} (-xy) + 3(2)(-xy)^{2} - (-xy)^{3} \\\\8 + 12 xy + 6x^{2} y^{2} + x^{3} y^{3} \\\\ordenas\\\\ x^{3} y^{3} + 6x^{2} y^{2} + 12 xy + 8$

Explicación paso a paso: