Please share any inspiration on this question

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ItsSKC

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Please share any inspiration on this question

« on: July 24, 2008, 01:27:44 am »

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x³+ax+b=0 where u³+v³=-b and 3uv=-a
show that x=u+v is a solution of x³+ax+b=0

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mark_alec

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Re: Please share any inspiration on this question

« Reply #1 on: July 24, 2008, 01:37:02 am »

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$x^3-3uvx-u^3-v^3 = 0$
Then to show that $u+v$ is a solution, just substitute it in.

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plato

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Re: Please share any inspiration on this question

« Reply #2 on: July 30, 2008, 12:00:53 am »

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An alternative, yet similar method can cut out the need to expand the cubic $(u+v)^3$

$x^3-3uvx=u^3+v^3$
$x(x^2-3uv)=(u+v)(u^2-uv+v^2)$
$x(x^2-3uv)=(u+v)(u^2-uv+v^2+3uv-3uv)$
$x(x^2-3uv)=(u+v)(u^2+2uv+v^2-3uv)$
$x(x^2-3uv)=(u+v)((u+v)^2-3uv)$

By equating the factors on the left and right sides, you can see that $x = u+v$

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Ahmad

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Re: Please share any inspiration on this question

« Reply #3 on: July 30, 2008, 03:35:49 am »

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Exploit the identity,

$a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - ac - bc)$

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