Find the value of $x^{3} - 6 x^{2} + 6 x$ $x^3-6x^2+6x$ when $x = 2 + 2^{\frac{2}{3}} + 2^{\frac{1}{3}} ?$ $x=2+2^(2/3)+2^(1/3)?$

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Answer 1 · 2018-06-20T10:45:44

$2$ $2$

$x^{3} - 6 x^{2} + 6 x$ $x^3-6x^2+6x$

= $x^{3} - 6 x^{2} + 12 x - 8 - 6 x + 12 - 4$ $x^3-6x^2+12x-8-6x+12-4$

= ${(x - 2)}^{3} - 6 \cdot (x - 2) - 4$ $(x-2)^3-6*(x-2)-4$

After setting $y = x - 2$ $y=x-2$ , this equation became $y^{3} - 6 y - 4$ $y^3-6y-4$

Hence,

$y = 2^{\frac{2}{3}} + 2^{\frac{1}{3}}$ $y=2^(2/3)+2^(1/3)$

$y^{3} = 4 + 6 \cdot 2^{\frac{2}{3}} + 6 \cdot 2^{\frac{1}{3}} + 2$ $y^3=4+6*2^(2/3)+6*2^(1/3)+2$

$y^{3} = 6 + 6 \cdot (2^{\frac{2}{3}} + 2^{\frac{1}{3}})$ $y^3=6+6*(2^(2/3)+2^(1/3))$

$y^{3} = 6 y + 6$ $y^3=6y+6$

$y^{3} - 6 y - 4 = 2$ $y^3-6y-4=2$

Find the value of x3−6x2+6xx^3-6x^2+6x when x=2+223+213? x=2+2^(2/3)+2^(1/3)?