Respuesta :
Explanation:
Knowing that:
[tex]\begin{gathered} x=\sqrt[3]{p+q}+\sqrt[3]{p-q} \\ r^3=p^2-q^2\text{ }\Rightarrow r=\sqrt[3]{p^2-q^2} \\ \end{gathered}[/tex]We have to prove that:
[tex]x^3-3rx-2p=0[/tex]First, let's find x³:
It is known that:
[tex](a+b)^3=a^3+b^3+3a^2b+3ab^2[/tex]So,
[tex]\begin{gathered} x^3=(\sqrt[3]{p+q}+\sqrt[3]{p-q})^3 \\ =\sqrt[3]{p+q}^3+\sqrt[3]{p-q}^3+3\sqrt[3]{p+q}^2\cdot\sqrt[3]{p-q}+3\sqrt[3]{p+q}\cdot\sqrt[3]{p-q}^2 \\ =p+q+p-q+3(p+q)^{\frac{2}{3}}\cdot\sqrt[3]{p-q}+3\sqrt[3]{p+q}\cdot(p-q)^{\frac{2}{3}} \\ =2p+3(p+q)^{\frac{2}{3}}\cdot\sqrt[3]{p-q}+3\sqrt[3]{p+q}\cdot(p-q)^{\frac{2}{3}} \\ =2p+3\sqrt[3]{(p+q)^2\cdot(p-q)}+3\sqrt[3]{(p-q)^2\cdot(p+q)} \\ =2p+3\sqrt[3]{(p^2+2pq+q^2)^{}\cdot(p-q)}+3\sqrt[3]{(p^2-2pq+q^2)^{}\cdot(p+q)} \\ =2p+3\sqrt[3]{p^3-p^2q+2p^2q-2pq^2+pq^2-q^3}+3\sqrt[3]{p^3-p^2q-2p^2q-2pq^2+2pq^2+q^3} \\ =2p+3\sqrt[3]{p^3+p^2q-pq^2-q^3}+3\sqrt[3]{p^3-p^2q-pq^2+q^3} \end{gathered}[/tex]Now, let's 3rx:
[tex]\begin{gathered} 3rx=3\cdot\sqrt[3]{p^2-q^2}\cdot(\sqrt[3]{p+q}+\sqrt[3]{p-q}) \\ =3\cdot\sqrt[3]{(p^2-q^2)\cdot(p+q)}+3\sqrt[3]{(p^2-q^2)\cdot(p-q)} \\ =3\sqrt[3]{p^3+p^2q-pq^2-q^3}+3\sqrt[3]{p^3-p^2q-pq^2+q^3} \end{gathered}[/tex]If we compare
[tex]\begin{gathered} x^3-3rx-2p \\ \end{gathered}[/tex]We have:
[tex]=2p+3\sqrt[3]{p^3+p^2q-pq^2-q^3}+3\sqrt[3]{p^3-p^2q-pq^2+q^3}-(3\sqrt[3]{p^3+p^2q-pq^2-q^3}+3\sqrt[3]{p^3-p^2q-pq^2+q^3})-2p[/tex]Subtracting the terms, we have:
[tex]x^3-3rx-2p=0[/tex]