@Adri van der Meer 

I would like to thank you for your help. That was exactly what I meant.

I am very sorry for unclear question.

I would like to thank you one more time,

Regards,

D.L.

@Carl Love

I have a question regardin multiplication sign.

I copy the formula from the resulted calucaltions of the maple.

It means that maple do not put the multipication sign in its solutions? Moreover, if I copy and paste the  polynom for my farther calcultions, does it mean that I am performing the calculation procedures in the wrong way ? 

I would like to thank you in advance.

Best regards,

D.L.

@Preben Alsholm 

I would like to thank you for your assistance.

@Carl Love 

I would like to thank you for your answer. Seems to be very helpful for me at the moment.

@Preben Alsholm  

I would like first of all to thank you for your help.

I have a very long term :

(2+k[1]*lambda^2*k[1](lambda*alpha[2]*k[2]+alpha[1](-2-2*k[2]+k[2]*lambda^2))*lambda(lambda*alpha[2]*k[2]+alpha[1](-2-2*k[2]+k[2]*lambda^2))^2-4*lambda*alpha[2]*k[1]*k[2]-lambda^5*alpha[2]*k[1]*k[2]+alpha[1]*k[1]*lambda^4*k[2]+lambda*alpha[2]*k[1]*lambda(2+5*k[2])^2+2*lambda+4*alpha[1]+2*k[1]-4*lambda*alpha[2]*k[2]-4*lambda*alpha[2]+k[1](lambda*alpha[2]*k[2]+alpha[1](-2-2*k[2]+k[2]*lambda^2))-4*lambda*alpha[2]*k[1]-k[1](lambda*alpha[2]*k[2]+alpha[1](-2-2*k[2]+k[2]*lambda^2))*lambda(lambda*alpha[2]*k[2]+alpha[1](-2-2*k[2]+k[2]*lambda^2))^2*k[1]-k[1]*lambda^2*k[1](lambda*alpha[2]*k[2]+alpha[1](-2-2*k[2]+k[2]*lambda^2))-lambda*k[2](k[1]*lambda*alpha[1]-2*k[1]*alpha[2]+k[1]*alpha[2]*lambda^2-2*alpha[2])*lambda(k[1]*lambda*alpha[1]-2*k[1]*alpha[2]+k[1]*alpha[2]*lambda^2-2*alpha[2])^2-alpha[1]*k[1]*lambda(2+5*k[2])^2+4*alpha[1]*k[1]*k[2]-2*alpha[1]*k[2]*lambda^2+2*lambda^3*alpha[2]*k[2]-2*k[1]*lambda^2+k[1](lambda*alpha[2]*k[2]+alpha[1](-2-2*k[2]+k[2]*lambda^2))*k[1]+lambda*k[2](k[1]*lambda*alpha[1]-2*k[1]*alpha[2]+k[1]*alpha[2]*lambda^2-2*alpha[2])-k[1](lambda*alpha[2]*k[2]+alpha[1](-2-2*k[2]+k[2]*lambda^2))*lambda(lambda*alpha[2]*k[2]+alpha[1](-2-2*k[2]+k[2]*lambda^2))^2+4*alpha[1]*k[2]+4*alpha[1]*k[1])*x*(4*alpha[1]+4*alpha[1]*k[2]-2*alpha[1]*k[2]*lambda^2+alpha[1]*k[1]*lambda^4*k[2]+4*alpha[1]*k[1]+4*alpha[1]*k[1]*k[2]-alpha[1]*k[1]*lambda(2+5*k[2])^2-4*lambda*alpha[2]-4*lambda*alpha[2]*k[2]+2*lambda^3*alpha[2]*k[2]-lambda^5*alpha[2]*k[1]*k[2]-4*lambda*alpha[2]*k[1]-4*lambda*alpha[2]*k[1]*k[2]+lambda*alpha[2]*k[1]*lambda(2+5*k[2])^2+2+k[1](lambda*alpha[2]*k[2]+alpha[1](-2-2*k[2]+k[2]*lambda^2))-k[1](lambda*alpha[2]*k[2]+alpha[1](-2-2*k[2]+k[2]*lambda^2))*lambda(lambda*alpha[2]*k[2]+alpha[1](-2-2*k[2]+k[2]*lambda^2))^2+2*lambda+lambda*k[2](k[1]*lambda*alpha[1]-2*k[1]*alpha[2]+k[1]*alpha[2]*lambda^2-2*alpha[2])-lambda*k[2](k[1]*lambda*alpha[1]-2*k[1]*alpha[2]+k[1]*alpha[2]*lambda^2-2*alpha[2])*lambda(k[1]*lambda*alpha[1]-2*k[1]*alpha[2]+k[1]*alpha[2]*lambda^2-2*alpha[2])^2)

and I pretend that this term can be rewriten in the following way:

x*k[1]^2*(lambda+1)^2*(lambda-1)^2*(1+(1-lambda^2)*k[1])(lambda*alpha[2]*k[2]+alpha[1]*((lambda^2-2)*k[2]-2))^2

Could you help me to find out the solution for my issue.

I would like to thank you in advance,

Regards,

D.L.

@Carl Love 

I would like to thank you for your reply.

When I use the above mentioned commands:

(expand@@2)~(expr1);
or
map(expand@@2, expr1); 

I still receive a messy term. Could you please write me down the steps, maybe I am wrong in programing procedures.

I would like to thank you in advance.
D.L.

@acer 

I would like to thank you. Now it looks beter. 

@Thomas Richard 

I would like to thnak you for your help frnd. 

@Minus one; @Markiyan Hirnyk 

I would like to thank you for your help.

I was expecting to obtain something similar to :

x*k[1]*(1+(1-lambda^2)*k[1])(lambda*alpha[2]*k[2]+alpha[1](-2+(-2+lambda^2)*k[2]))^2

now, I am totaly lost... Is it possible at least to factor out something similar ?

I would like to thank you in advance.

D.L.

@gkokovidis 

I would like to thnak you for you reply ! it was very helpuful for me ! I did not write in clear manner the formulas, nevertheless I took into consideration your way of solving. It a good point for me for an alternative way of solving.

I would like also to ask you two more question:

1) Is it possible to see the steps how Maple 15 solve this system of equation ?

2) The denominators of both results are equal to: (-5*k[2]*k[1]*lambda^2-2*k[2]*lambda^2-2*k[1]*lambda^2+4*k[1]+4*k[2]*k[1]+4+k[2]*lambda^4*k[1]+4*k[2]).

In the original solutions one have written the denominator in the following way : [4-2(-2-lambda^2)k[2]+k[1](k[2] lambda^4 +4(1+k[2] )-lambda^(2)(2+5k[2] )) ]^(2).

Everything is clear for me untill to the moment where it comes from the power "two". Is it possible to rearrange the terms in such way that denominator stands in power two ?

I would like to thank you in advance for your time and consideration. I am looking forward for you reply.

Best regrads,

D.L.

@Carl Love 

I would like to thnak you for you reply ! it was very helpuful for me ! Nevertheless, I have 2 more questions.

I am sorry for bothering you, I will be very glad for you assistane and help.

1) Is it possible to see the steps how Maple 15 solve this system of equation ?

2) The denominators of both results are equal to: (-5*k[2]*k[1]*lambda^2-2*k[2]*lambda^2-2*k[1]*lambda^2+4*k[1]+4*k[2]*k[1]+4+k[2]*lambda^4*k[1]+4*k[2]).

In the original solutions one have written the denominator in the following way : [4-2(-2-lambda^2)k[2]+k[1](k[2] lambda^4 +4(1+k[2] )-lambda^(2)(2+5k[2] ) )]^(2).

Everything is clear for me untill to the moment where it comes from the power "two". Is it possible to rearrange the terms in such way that denominator stands in power two ?

I would like to thank you in advance for your time and consideration. I am looking forward for you reply.

Best regrads,

D.L.