If x and Y are the roots of equation x^2+4x-4=0 , 1/(x-2)^3 + 1/(y-2)^3=?

can we put z instead of x??

ok

(x-z)(x-y) = x^2+4x-4 zy = 16/2 = -4 z+y = -4

-5/8

\[ \frac{(y-2)^3 +(z-2)^3}{((y-2)(z-2))^3} = \frac{(y+z-4)((y-2)^2 - (y-2)(z-2) + (z-2)^2 }{(yz - 2 (y+z) + 4)^3}\]

X=2 sqrt(2)-2 Y=-2 sqrt(2)-2 REPLACE THEM THERE U GET 1/((-2)sqrt(2)-2-2)^3+1/(2sqrt(2)-2-2)^3=-0.625

I think we need to evaluate y^2+z^2 \[ \left ( \frac{-b + \sqrt{b^2 - 4ac}}{2a} \right )^2 + \left ( \frac{-b - \sqrt{b^2 - 4ac}}{2a} \right )^2 = \frac{2b^2 + 2(b^2 - 4ac)}{4a^2} \\ \\ = \frac{4b^2 - 8ac}{4a^2} = \frac{4.4^2 - 8.1.4}{4} \] evaluate rest ... and put values ... and there might me some errors ... please check yourself

@experimentX umissunderstood the question i think ! open study did u found the roots ! like those X=2 sqrt(2)-2 Y=-2 sqrt(2)-2 !?

well ... i believe this is more general way of solving problem unless you are still junior at secondary school

no!

yeh sorry any way we trying to solve this question now right ^_^

x+y=-4 xy=-4

keep trying ... :)

@experimentX is that the first step

yes!! you can find it by quadratic formula ... sum of roots and product of roots!! but verify!! I'm a bit inaccurate in calculations!!

thanzz a lot @Callisto @experimentX @badi

Don't just copy the answer!!!! Do it on you own!!!!

ok

thanzz for giving me an idea

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