Question

Untuk [tex]\displaystyle x\in \mathbb{R}[/tex] carilah nilai dari
[tex]\displaystyle \sqrt[3]{\frac{27+11\sqrt{6}}{72}}+\sqrt[3]{\frac{27-11\sqrt{6}}{72}}[/tex]

Answer 1

Verified answer

Jawab:

[tex]\sqrt[3]{\dfrac{27+11\sqrt{6}}{72}}=\dfrac{1}{2}\sqrt[3]{3+\frac{11}{9}\sqrt{6}}\\\sqrt[3]{\dfrac{27-11\sqrt{6}}{72}}=\dfrac{1}{2}\sqrt[3]{3-\frac{11}{9}\sqrt{6}}\\S=\sqrt[3]{\dfrac{27+11\sqrt{6}}{72}}+\sqrt[3]{\dfrac{27-11\sqrt{6}}{72}}=\dfrac{1}{2}(\sqrt[3]{3+\frac{11}{9}\sqrt{6}}+\sqrt[3]{3-\frac{11}{9}\sqrt{6}})\\\text{Misal :} x=\sqrt[3]{3+\frac{11}{9}\sqrt{6}} \text{ dan } y=\sqrt[3]{3-\frac{11}{9}\sqrt{6}}[/tex]

[tex]({a+b\sqrt{c}})^3=a^3+3a^2b\sqrt{c}+3a(b\sqrt{c})^2+(b\sqrt{c})^3=a^3+3ab^2c+(3a^2b+b^3c)\sqrt{c}\\c=6\rightarrow a^3+18ab^2=3, 3a^2b+6b^3=\frac{11}{9}\\a=1, \rightarrow b=\frac{1}{3} \qquad(M)\\a=2,\rightarrow b=\sqrt{\frac{-5}{36}} \qquad (TM)\\\therefore x=1+\frac{1}{3}\sqrt{6},y=1-\frac{1}{3}\sqrt{6}\\[/tex]

[tex]S = \dfrac{1}{2}(x +y) = \dfrac{1}{2}(2)=1[/tex]