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Algebra Level 4

If x = − 2020 x=-2020 . What can we say about the number y 3 + y 2 + y , y^3+y^2+y, where

y = 1 6 ( 2 2 3 3 81 x 2 − 42 x + 9 − 27 x + 7 3 − 4 2 3 3 81 x 2 − 42 x + 9 − 27 x + 7 3 − 2 ) ? \small y=\frac{1}{6}\left(2^{\frac{2}{3}}\sqrt[3]{3\sqrt{81x^2-42x+9}-27x+7} -\frac{4\sqrt[3]{2}}{\sqrt[3]{3\sqrt{81x^2-42x+9}-27x+7}}-2\right)?

Note Try to do it without using calculators.

y 3 + y 2 + y y^3+y^2+y is real and in the interval [ 0 , 1000 ) [0, 1000) y 3 + y 2 + y y^3+y^2+y is real and in the interval [ 3000 , ∞ ) [3000, \infty) y 3 + y 2 + y y^3+y^2+y is a non real complex number y 3 + y 2 + y y^3+y^2+y is real and in the interval [ 1000 , 2000 ) [1000, 2000) y 3 + y 2 + y y^3+y^2+y is real and y < 0 y<0 y 3 + y 2 + y y^3+y^2+y is real and in the interval [ 2000 , 3000 ) [2000, 3000)

Arturo Presa by Arturo Presa , 62, USA

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2 solutions

Arturo Presa
Dec 13, 2020

Define u = 3 81 x 2 − 42 x + 9 3 u= 3\sqrt[3]{81x^2-42x+9} and v = 27 x − 7. v=27x-7. It is easy to see that ( u − v ) ( u + v ) = 32. ( ∗ ) (u-v)(u+v)=32. (*)

Then we can write the number y y in the form y = 1 6 ( 2 2 / 3 u − v 3 − 4 2 3 u − v 3 − 2 ) . y=\frac{1}{6}(2^{2/3} \sqrt[3]{u-v}-\frac{4\sqrt[3]{2}}{\sqrt[3]{u-v}}-2). Then y + 1 3 = 1 6 ( 2 2 / 3 u − v 3 − 4 2 3 u − v 3 ) . y+\frac{1}{3}=\frac{1}{6}(2^{2/3} \sqrt[3]{u-v}-\frac{4\sqrt[3]{2}}{\sqrt[3]{u-v}}). Multiplying the term 4 2 3 u − v 3 \large\frac{4\sqrt[3]{2}}{\sqrt[3]{u-v}} by u + v 3 u + v 3 , \large\frac{\sqrt[3]{u+v}}{\sqrt[3]{u+v}}, and using the property (*) to simplify we get y + 1 3 = 1 6 ( 2 2 / 3 u − v 3 − 2 2 / 3 u + v 3 ) = 2 2 / 3 6 ( u − v 3 − u + v 3 ) y+\frac{1}{3}=\frac{1}{6}(2^{2/3} \sqrt[3]{u-v}- 2^{2/3}{\sqrt[3]{u+v}})=\frac{2^{2/3}}{6} (\sqrt[3]{u-v}- {\sqrt[3]{u+v}}) Raise both sides of the last equation to the third power and simplify. Then you get ( y + 1 3 ) 3 = − x + 7 27 − 2 3 ( y + 1 3 ) (y+\frac{1}{3})^3=-x+\frac{7}{27}-\frac{2}{3}(y+\frac{1}{3}) Or, equivalently ( y + 1 3 ) 3 + 2 3 ( y + 1 3 ) = − x + 7 27 (y+\frac{1}{3})^3+\frac{2}{3}(y+\frac{1}{3})=-x+\frac{7}{27} Expanding the left side and combining like terms we get, y 3 + y 2 + y + 7 27 = − x + 7 27 y^3+y^2+y+\frac{7}{27}=-x+\frac{7}{27} Hence, y 3 + y 2 + y = − x = 2020 y^3+y^2+y=-x=2020 and, therefore, y 3 + y 2 + y y^3+y^2+y is a real number in the interval [ 2000 , 3000 ) . [2000, 3000).

1 Helpful 1 Interesting 3 Brilliant 0 Confused
Pi Han Goh
Dec 14, 2020

Let z = 3 81 x 2 − 42 x + 9 − 27 x + 7 z = 3\sqrt{81x^2 - 42x + 9} - 27x + 7 . Rearranging gives ( z + 27 x − 7 ) 2 = 9 ( 81 x 2 − 4 x + 9 ) ( z+27x - 7)^2= 9 (81x^2 - 4x + 9 ) . Solving for x x gives x = − z 2 + 14 z + 32 54 z . x = \dfrac{-z^2 + 14z + 32}{54z} . With x = − 2020 x = -2020 , we have − 2020 = − z 2 + 14 z + 32 54 z ⟹ z 2 − 109094 z − 32 = 0 ⟹ z − 32 z = 109094 -2020 = \dfrac{-z^2 + 14z + 32}{54z} \quad\implies \quad z^2 - 109094 z - 32 = 0 \quad \implies \quad z - \frac {32}z = 109094 On the other hand, we can express y y in terms of z z , y = 1 6 ( 2 2 3 z 3 − 4 ⋅ 2 3 z 3 − 2 ) ⟹ 6 y + 2 4 3 = z 3 − 32 3 z 3 y = \frac16 \left ( 2^{\frac23} \sqrt[3]{z} - \frac{ 4\cdot \sqrt[3]{2} }{\sqrt[3]{z}} - 2 \right)\quad \implies \quad \dfrac{6y+2}{\sqrt[3]{4}} = \sqrt[3]{z} - \dfrac{\sqrt[3]{32}}{\sqrt[3]{z}} Cube both sides gives ( 6 y + 2 ) 3 4 = ( z − 32 z ) − 3 ⋅ ( 6 y + 2 ) 4 3 ⋅ 32 3 0 ( 6 y + 2 ) 3 4 = 109094 − 6 ( 6 y + 2 ) 0 54 y 3 + 54 y 2 + 54 y − 109080 = 0 0 y 3 + y 2 + y = 109080 54 = 2020 \begin{array} {r c l} \dfrac{(6y + 2)^3}4 &=& \left( z - \dfrac{32}z \right) - 3 \cdot \dfrac{(6y + 2)}{\sqrt[3]{4}} \cdot \sqrt[3]{32} \\ \phantom0\\ \dfrac{(6y+2)^3}4 &=& 109094 - 6(6y + 2) \\ \phantom0\\ 54y^3 + 54y^2 + 54y - 109080 &=& 0 \\ \phantom0\\ y^3 + y^2 + y &=& \dfrac{109080}{54} = \boxed{2020} \end{array}

Likewise, we can generalize this to:

If y = 1 6 ( 2 2 3 3 81 x 2 − 42 x + 9 − 27 x + 7 3 − 4 2 3 3 81 x 2 − 42 x + 9 − 27 x + 7 3 − 2 ) y=\frac{1}{6}\left(2^{\frac{2}{3}}\sqrt[3]{3\sqrt{81x^2-42x+9}-27x+7} -\frac{4\sqrt[3]{2}}{\sqrt[3]{3\sqrt{81x^2-42x+9}-27x+7}}-2\right) then y 3 + y 2 + y = − x . y^3 + y^2 + y = -x .

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