If-x-2-2-2-3-2-1-3-then-prove-that-x-3-6x-2-6x-2-0-

Question Number 173152 by AgniMath last updated on 07/Jul/22

If x = 2 + 2^{\frac{2}{3}} + 2^{\frac{1}{3}} then prove that

x^{3} - 6 x^{2} + 6 x - 2 = 0 .

Answered by Frix last updated on 07/Jul/22

let x = y^{3} + y^{2} + y

y^{9} + 3 y^{8} + 6 y^{7} + y^{6} - 6 y^{5} - 15 y^{4} - 5 y^{3} + 6 y - 2 = 0

y = 2^{1 / 3}

8 + 2^{2 / 3} 12 + 2^{1 / 3} 24 + 4 - 2^{2 / 3} 12 - 2^{1 / 3} 30 - 10 + 2^{1 / 3} 6 - 2 = 0

{ir}^{'} s easy to see this is true .

Answered by som(math1967) last updated on 07/Jul/22

x - 2 = 2^{\frac{2}{3}} + 2^{\frac{1}{3}}

{(x - 2)}^{3} = {(2^{\frac{2}{3}} + 2^{\frac{1}{3}})}^{3}

\Rightarrow x^{3} - 6 x^{2} + 12 x - 8 = {(2^{\frac{2}{3}})}^{3} + {(2^{\frac{1}{3}})}^{3}

+ 3 . 2^{\frac{2}{3}} . 2^{\frac{1}{3}} (2^{\frac{2}{3}} + 2^{\frac{1}{3}})

\Rightarrow x^{3} - 6 x^{2} + 12 x - 8 = 4 + 2 + 3.2 (x - 2)

[∵ (x - 2) = 2^{\frac{2}{3}} + 2^{\frac{1}{3}}]

\Rightarrow x^{3} - 6 x^{2} + 12 x - 8 = 6 + 6 x - 12

\Rightarrow x^{3} - 6 x^{2} + 12 x - 6 x - 8 + 6 = 0

∴ x^{3} - 6 x^{2} + 6 x - 2 = 0

Commented by Tawa11 last updated on 11/Jul/22

Great sir

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