A simple answer to a complicated x x

Algebra Level 3

x = 7 + 22 3 + 7 22 3 x = \sqrt[3]{7+\sqrt{22}}+ \sqrt[3]{7- \sqrt{22}} . In the equation x 3 + b x + c = 0 x^3 + bx + c = 0 , what is b c -bc if b b and c c are integers?


The answer is -126.

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3 solutions

Let α = 7 + 22 3 \alpha = \sqrt[3]{7+\sqrt{22}} and β = 7 22 3 \beta = \sqrt[3]{7-\sqrt{22}} , then x = α + β x = \alpha+\beta .

x 3 + b x + c = 0 α 3 + 3 α 2 β + 3 α β 2 + β 3 + b x + c = 0 7 + 22 + 3 α β ( α + β ) + 7 22 + b x + c = 0 14 + 3 α β x + b x + c = 0 \begin{aligned} x^3 + bx + c & = 0 \\ \alpha^3 + 3\alpha^2 \beta + 3\alpha \beta^2 + \beta^3 + bx + c & = 0 \\ 7+\sqrt{22} + 3\alpha \beta ( \alpha + \beta) + 7- \sqrt{22} + bx +c & = 0 \\ \color{#3D99F6}{14} + \color{#D61F06}{3\alpha \beta} x + \color{#D61F06}{b}x + \color{#3D99F6}{c} & = 0 \end{aligned}

{ b = 3 α β = 3 ( 7 + 22 3 ) ( 7 22 3 ) = 3 49 22 3 = 3 27 3 = 9 c = 14 \Rightarrow \begin{cases} b & = -3\alpha \beta = -3 \left(\sqrt[3]{7+\sqrt{22}}\right) \left( \sqrt[3]{7-\sqrt{22}}\right) \\ & = -3 \sqrt[3]{49-22} = -3 \sqrt[3]{27} = -9 \\ \\ c & = -14 \end{cases}

b c = ( 9 ) ( 14 ) = 126 \Rightarrow -bc = - (-9)(-14) = \boxed{-126} .

Cube both sides:

x 3 = 7 + 22 + 3 ( 7 + 22 ) ( 7 22 ) 3 ( 7 + 22 3 + 7 22 3 ) + 7 22 x 3 = 14 + 3 49 22 3 x x 3 = 14 + 9 x x 3 9 x 14 = 0 b = 9 c = 14 b c = 126 x^3=7+\sqrt{22}+3\sqrt[3]{(7+\sqrt{22})(7-\sqrt{22})}(\sqrt[3]{7+\sqrt{22}}+\sqrt[3]{7-\sqrt{22}})+7-\sqrt{22} \\ x^3=14+3\sqrt[3]{49-22}x \\ x^3=14+9x \\ x^3-9x-14=0 \\ \implies b=-9 \implies c=-14 \implies -bc=\boxed{-126}

Jonathan Hsu
Jul 17, 2015

We can set x x as y + z y + z . We can easily see that y z yz = 3 3 and y 3 + z 3 = 14 y^3 + z^3 = 14 . Expanding x 3 x^3 gives ( y + z ) 3 (y+z)^3 which is ( y 3 + z 3 + 3 ( y 2 z + y z 2 ) ) = ( ( y 3 + z 3 + 3 y z ( z + y ) ) (y^3 + z^3 + 3(y^2*z + yz^2)) = ((y^3 + z^3 + 3yz(z + y)) . Now we can substitute values, which gives us 14 + 9 ( y + z ) 14 + 9(y+z) . The whole expression is now: 14 + 9 ( y + z ) + b ( y + z ) + c 14 + 9(y+z) + b(y+z) + c . Clearly, we can see to cancel out the terms, b = 9 b = 9 and c = 14 c = 14 . So, b c = 126 b*c = -126 .

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