IIT JEE , AIEEE Forum - Mathematics , Trignometry

karthik_18049200

please solve this experts (1+2x)(1+2y)=3,then prove that
1+8x³/x(1-x²)=1+8y³/y(1-y³)

bvsatyaram

Once double check your question...
Because, the x^3 in the numerator and x in the denominator cancel....

karthik_18049200

no

the numerator is 1+8x³ and denominator is x(1-x³)

similar for y also

raulrag009

(1+2x)(1+2y)=3

thus x=(1-y)/1+2y

LHS = (1+8x³)/x(1-x³)

= 1+8[ (1-y)/(1+2y)]³ whole divided by [(1-y)/1+2y] [1 - {(1-y)/1+2y}³]

= (1+2y)³ + 8(1-y)³ whole divided by [(1-y)/1+2y] [(1+2y⁾³ - (1-y^)3]

As (1+2y)³ gets cancelled in numerator and denominator after takin LCM

= expand using (a+b)³ = a³ + b³ +3a²b + 3ab²

= [1+8y³ +6y+12y² +8(1-y³-3y+3y²)](1+2y) whole divided by

(1-y)[1+8y³+6y+12y² -1+y³+3y-3y²]

= 9[4y²-2y+1](1+2y) whole divided by (1-y)9(y³+y²+y)

= [4y²-2y+1](1+2y) / y(1-y)(y²+y+1)

1+8y^3 =(1+2y)[1+4y²-2y] and 1-y³=(1-y)(1+y+y²)

= 1+8y³ / y(1-y³)

=RHS

konichiwa2x

this question interests me. There must surely be an easy method to prove the result...Do you have any leads? (you have posted it in the trig section)

konichiwa2x

we have

Let

now

,

and

.

Thus

The problem is asking you to show that

, which is very easy to verify!

iitkgp_bipin

Good work guys.

hsbhatt

Yeah, this is good stuff. Makes me fall more and more in love with mathematics

hsbhatt

I just wanted to see how the pros do it, so I put up this problem in a forum and the soln came in 27 mins flat:

Let $a = x + y, b = xy,$ we have $(1 + 2x)(1 + 2y) = 3Longleftrightarrow a + 2b = 1.$

$herefore x(1 - x^3)(1 + 8y^3) - y(1 - y^3)(1 + 8x^3)$

$= (x - y){1 - 8xy(x + y) - (x + y)(x^2 + y^2) - 8x^3y^3}$

$=(x-y){1-ab-a(a^2-2b)-8b^3}$

$= (y - x)(a^3 + 8b^3 - 1 + 6ab)$

$= (y - x)(a + 2b - 1)(a^2 - 2ab + 4b^2 + 1 - 2ab + a + 2b) = 0.$ Q.E.D.

These guys work magic with algebraic manipulations!

konichiwa2x

cool!

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