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Question-215092




Question Number 215092 by BaliramKumar last updated on 28/Dec/24
Commented by BaliramKumar last updated on 28/Dec/24
I want to find the value of a and b
Iwanttofindthevalueofaandb
Commented by BaliramKumar last updated on 28/Dec/24
a^3 −b^3  = 125  a^2 −b^2  = 25  a+b = 5  a^2  + ab + b^(2 )  = ?
a3b3=125a2b2=25a+b=5a2+ab+b2=?
Answered by Rasheed.Sindhi last updated on 28/Dec/24
a^2 −b^2 =25⇒(a−b)(a+b)=25  (a−b)(5)=25⇒a−b=5   { ((a+b=5)),((a−b=5)) :}⇒a=5,b=0  a^2 +ab+b^2 =5^2 +5×0+0^2 =25
a2b2=25(ab)(a+b)=25(ab)(5)=25ab=5{a+b=5ab=5a=5,b=0a2+ab+b2=52+5×0+02=25
Commented by A5T last updated on 28/Dec/24
We can also do this,so we get a contradiction.  a^3 −b^3 =(a−b)(a^2 +ab+b^2 )=625  a^2 −b^2 =(a−b)(a+b)=25  a+b=5⇒a−b=((25)/5)=5  ⇒a^2 +ab+b^2 =((625)/5)=125
Wecanalsodothis,sowegetacontradiction.a3b3=(ab)(a2+ab+b2)=625a2b2=(ab)(a+b)=25a+b=5ab=255=5a2+ab+b2=6255=125
Commented by Frix last updated on 28/Dec/24
We′ve got 3 equations for 2 unknowns and  all must be satisfied.  (1)     b=5−a  (2)     a^2 =25+b^2        a^2 =25+(5−a)^2        a^2 =50−10a+a^2  ⇒ a=5 ⇒ b=0  (3)     Test a^3 −b^3 =625 ⇒ wrong  ⇒  No solution
Wevegot3equationsfor2unknownsandallmustbesatisfied.(1)b=5a(2)a2=25+b2a2=25+(5a)2a2=5010a+a2a=5b=0(3)Testa3b3=625wrongNosolution
Commented by BaliramKumar last updated on 29/Dec/24
i think question is wrong
ithinkquestioniswrong
Commented by Rasheed.Sindhi last updated on 29/Dec/24
I think that the question has come   before this time also.
Ithinkthatthequestionhascomebeforethistimealso.
Commented by A5T last updated on 29/Dec/24
Q211393
Q211393
Commented by Frix last updated on 29/Dec/24
A question can not be wrong.  The proposed answers are wrong.
Aquestioncannotbewrong.Theproposedanswersarewrong.

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