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# A Problem Of IMO

If x = $$2 + \sqrt[3]{2} + \sqrt[3]{2^2}$$, Then $$x^3 - 6x^2 + 6x$$ = ?

Note by Swapnil Das
1 year, 7 months ago

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Take 2 on LHS and cube on both the sides

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

Simplifing

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

Rearranging (formation of x on RHS)

$$\displaystyle x^3 -6x^2 + 12x -8 = 6x -6$$

$$\displaystyle x^3 - 6x^2 + 6x = 2$$ · 1 year, 7 months ago

Can U please recommend me some books for trigonometry? · 1 year, 7 months ago

I never read any Trigonometry books but I am sure Omkar kulkarni can help you out(I am unable to tag him right now) · 1 year, 7 months ago

Thanks for suggestion! · 1 year, 7 months ago

Thank You. You were really helpful! · 1 year, 7 months ago