IIT JEE , AIEEE Forum - Mathematics , Algebra - Cubic has its own mood- no one can hange it

Mr.IITIAN007

Hey buddiesssss.........try this one ;-
If alpha , beta , gamma are the roots of the cubic equarion

a x³ + b x² + c x +d = 0

Then what is the equation whose roots are
1+alpha / 1 - alpha , 1 + beta/ 1 - beta , 1+gamma /1 - gamma ?

Please guys...I don't know the answer...the one with perfect explanation will be rated.

spt_001

instead of x substitute (x-1)/(x+1)

Mr.IITIAN007

Hey wait a minute damned........have you tried to solve it....I think you haven't because replacing x by x-1 / x+1 is the actual technique of solving the problem but ...here it is forming a huge expression where just four variables are being cancelled(due to opposite sgns).
Its easy to say something but that much difficult to do it.

Mr.IITIAN007

oh come on ....Knock Knock.....o.k just tell me the answer at least I will definitely rate.

lazycol

change x to (x-1)/(x+1)

so a(x-1)³/(x+1)³+ b(x-1)²/(x+1)²+ c(x-1)/(x+1) + d = 0

a(x-1)³+ b(x-1)²(x+1) + c(x-1)(x+1)²+ d(x+1)³=0

(a+b+c+d)x³ - (3a+b-c-3d)x² + (3a-b-c+3d)x - (a-b+c-d) = 0

Mr.IITIAN007

Now that I call a perfect answer. Thanx Lazycol.

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