Resolver (sqrt{2}+1)^3-(sqrt{2}-1)^3

$$( \sqrt { 2 } + 1 ) ^ { 3 } - ( \sqrt { 2 } - 1 ) ^ { 3 }$$

Avaliar

$$14$$

Passos Para a Resolução

Utilize o teorema binomial \left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} para expandir \left(\sqrt{2}+1\right)^{3}.

$$\left(\sqrt{2}\right)^{3}+3\left(\sqrt{2}\right)^{2}+3\sqrt{2}+1-\left(\sqrt{2}-1\right)^{3}$$

O quadrado de \sqrt{2} é 2.

$$\left(\sqrt{2}\right)^{3}+3\times 2+3\sqrt{2}+1-\left(\sqrt{2}-1\right)^{3}$$

Multiplique 3 e 2 para obter 6.

$$\left(\sqrt{2}\right)^{3}+6+3\sqrt{2}+1-\left(\sqrt{2}-1\right)^{3}$$

Some 6 e 1 para obter 7.

$$\left(\sqrt{2}\right)^{3}+7+3\sqrt{2}-\left(\sqrt{2}-1\right)^{3}$$

Utilize o teorema binomial \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} para expandir \left(\sqrt{2}-1\right)^{3}.

$$\left(\sqrt{2}\right)^{3}+7+3\sqrt{2}-\left(\left(\sqrt{2}\right)^{3}-3\left(\sqrt{2}\right)^{2}+3\sqrt{2}-1\right)$$

O quadrado de \sqrt{2} é 2.

$$\left(\sqrt{2}\right)^{3}+7+3\sqrt{2}-\left(\left(\sqrt{2}\right)^{3}-3\times 2+3\sqrt{2}-1\right)$$

Multiplique -3 e 2 para obter -6.

$$\left(\sqrt{2}\right)^{3}+7+3\sqrt{2}-\left(\left(\sqrt{2}\right)^{3}-6+3\sqrt{2}-1\right)$$

Subtraia 1 de -6 para obter -7.

$$\left(\sqrt{2}\right)^{3}+7+3\sqrt{2}-\left(\left(\sqrt{2}\right)^{3}-7+3\sqrt{2}\right)$$

Para calcular o oposto de \left(\sqrt{2}\right)^{3}-7+3\sqrt{2}, calcule o oposto de cada termo.

$$\left(\sqrt{2}\right)^{3}+7+3\sqrt{2}-\left(\sqrt{2}\right)^{3}+7-3\sqrt{2}$$

Combine \left(\sqrt{2}\right)^{3} e -\left(\sqrt{2}\right)^{3} para obter 0.

$$7+3\sqrt{2}+7-3\sqrt{2}$$

Some 7 e 7 para obter 14.

$$14+3\sqrt{2}-3\sqrt{2}$$

Combine 3\sqrt{2} e -3\sqrt{2} para obter 0.

$$14$$

Fatorizar

$$2\times 7$$