Resolver (2a+1)^3-2a(2a+1)^2-(2a)^2-1

$$( 2 a + 1 ) ^ { 3 } - 2 a ( 2 a + 1 ) ^ { 2 } - ( 2 a ) ^ { 2 } - 1$$

Calcular

$$4a$$

Pasos da solución

Usar teorema binomial \left(p+q\right)^{3}=p^{3}+3p^{2}q+3pq^{2}+q^{3} para expandir \left(2a+1\right)^{3}.

$$8a^{3}+12a^{2}+6a+1-2a\left(2a+1\right)^{2}-\left(2a\right)^{2}-1$$

Usar teorema binomial \left(p+q\right)^{2}=p^{2}+2pq+q^{2} para expandir \left(2a+1\right)^{2}.

$$8a^{3}+12a^{2}+6a+1-2a\left(4a^{2}+4a+1\right)-\left(2a\right)^{2}-1$$

Expande \left(2a\right)^{2}.

$$8a^{3}+12a^{2}+6a+1-2a\left(4a^{2}+4a+1\right)-2^{2}a^{2}-1$$

Calcula 2 á potencia de 2 e obtén 4.

$$8a^{3}+12a^{2}+6a+1-2a\left(4a^{2}+4a+1\right)-4a^{2}-1$$

Usa a propiedade distributiva para multiplicar -2a por 4a^{2}+4a+1.

$$8a^{3}+12a^{2}+6a+1-8a^{3}-8a^{2}-2a-4a^{2}-1$$

Combina 8a^{3} e -8a^{3} para obter 0.

$$12a^{2}+6a+1-8a^{2}-2a-4a^{2}-1$$

Combina 12a^{2} e -8a^{2} para obter 4a^{2}.

$$4a^{2}+6a+1-2a-4a^{2}-1$$

Combina 6a e -2a para obter 4a.

$$4a^{2}+4a+1-4a^{2}-1$$

Combina 4a^{2} e -4a^{2} para obter 0.

$$4a+1-1$$

Resta 1 de 1 para obter 0.

$$4a$$

Expandir

$$4a$$

Pasos da solución

Usar teorema binomial \left(p+q\right)^{3}=p^{3}+3p^{2}q+3pq^{2}+q^{3} para expandir \left(2a+1\right)^{3}.

$$8a^{3}+12a^{2}+6a+1-2a\left(2a+1\right)^{2}-\left(2a\right)^{2}-1$$

Usar teorema binomial \left(p+q\right)^{2}=p^{2}+2pq+q^{2} para expandir \left(2a+1\right)^{2}.

$$8a^{3}+12a^{2}+6a+1-2a\left(4a^{2}+4a+1\right)-\left(2a\right)^{2}-1$$

Expande \left(2a\right)^{2}.

$$8a^{3}+12a^{2}+6a+1-2a\left(4a^{2}+4a+1\right)-2^{2}a^{2}-1$$

Calcula 2 á potencia de 2 e obtén 4.

$$8a^{3}+12a^{2}+6a+1-2a\left(4a^{2}+4a+1\right)-4a^{2}-1$$

Usa a propiedade distributiva para multiplicar -2a por 4a^{2}+4a+1.

$$8a^{3}+12a^{2}+6a+1-8a^{3}-8a^{2}-2a-4a^{2}-1$$

Combina 8a^{3} e -8a^{3} para obter 0.

$$12a^{2}+6a+1-8a^{2}-2a-4a^{2}-1$$

Combina 12a^{2} e -8a^{2} para obter 4a^{2}.

$$4a^{2}+6a+1-2a-4a^{2}-1$$

Combina 6a e -2a para obter 4a.

$$4a^{2}+4a+1-4a^{2}-1$$

Combina 4a^{2} e -4a^{2} para obter 0.

$$4a+1-1$$

Resta 1 de 1 para obter 0.

$$4a$$