A Problem Of IMO

Please help me.

If x = 2+23+2232 + \sqrt[3]{2} + \sqrt[3]{2^2}, Then x36x2+6xx^3 - 6x^2 + 6x = ?

#Algebra #IMO #Problems

Note by Swapnil Das
6 years, 2 months ago

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Comments

Take 2 on LHS and cube on both the sides

(x2)3=2(1+213)3\displaystyle (x-2)^3 = 2(1 + 2^{\frac{1}{3}})^3

Simplifing

x36x2+12x8=6(1+213+223)\displaystyle x^3 -6x^2 + 12x - 8 = 6(1 + 2^{\frac{1}{3}} + 2^{\frac{2}{3}})

Rearranging (formation of x on RHS)

x36x2+12x8=6x6\displaystyle x^3 -6x^2 + 12x -8 = 6x -6

x36x2+6x=2\displaystyle x^3 - 6x^2 + 6x = 2

Krishna Sharma - 6 years, 2 months ago

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Thank You. You were really helpful!

Swapnil Das - 6 years, 2 months ago

Can U please recommend me some books for trigonometry?

Swapnil Das - 6 years, 2 months ago

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I never read any Trigonometry books but I am sure Omkar kulkarni can help you out(I am unable to tag him right now)

Krishna Sharma - 6 years, 2 months ago

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@Krishna Sharma Thanks for suggestion!

Swapnil Das - 6 years, 2 months ago

x=2

Jacob Johanssen - 6 years, 1 month ago
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