Estude! Com Prof. Caju

Se (k + 1/k)^2 = 3, calcule k³ + 1/k³

bê1,

[tex3](k+\frac{1}{k})^3 = k^3+\frac{1}{k^3}+3(k+\frac{1}{k})\\
k^3+\frac{1}{k^3} = (k+\frac{1}{k})^3-3(k+\frac{1}{k})(I)\\
(k+\frac{1}{k})^2 =3 \implies (k+\frac{1}{k})=\pm\sqrt3(II)\\
(II)em(I): k^3+\frac{1}{k^3} = (\sqrt3)^3-3(\sqrt3)(I)=3\sqrt3-3\sqrt3=0\\
k^3+\frac{1}{k^3} = (-\sqrt3)^3-3(-\sqrt3)(I)=-3\sqrt3+3\sqrt3=0\\

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Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3](k+\frac{1}{k})^3 = k^3+\frac{1}{k^3}+3(k+\frac{1}{k})\\
k^3+\frac{1}{k^3} = (k+\frac{1}{k})^3-3(k+\frac{1}{k})(I)\\
(k+\frac{1}{k})^2 =3 \implies (k+\frac{1}{k})=\pm\sqrt3(II)\\
(II)em(I): k^3+\frac{1}{k^3} = (\sqrt3)^3-3(\sqrt3)(I)=3\sqrt3-3\sqrt3=0\\
k^3+\frac{1}{k^3} = (-\sqrt3)^3-3(-\sqrt3)(I)=-3\sqrt3+3\sqrt3=0\\

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