prove the following.?

Question

[a^(m)+b^(2)][a^(n)+y^(2)]
=[a^{(m+n)/2}+b*y]^(2)
+[y*{a^(m/2)} - {a^(n/2)}*b]^(2)

hint :- use lagranges identity.

lagranges identity:-
[a^(2)+b^(2)][x^(2)+y^(2)]=[a*x+b*y]^(2)+[a*y-b*x]^(2)

lagranges identity:-
[a^(2)+b^(2)][x^2+y^(2)}=
[a*x+b*y]^(2)+[a*y-b*x]^(2)

Answer 1

First of all, u do not need Lagrange's identity to prove it. u just need to expand everything.

So, u want to use Lagrange's identity, fine. set a^m=a^2, a^n=x^2, substitute these into the Lagrange's identity and u get exactly [a^{(m+n)/2}+b*y]^(2) + [y*{a^(m/2)} - {a^(n/2)}*b]^(2) without even needing to do anything else.

This is not called proving!

Answer 2

We get it by expansion but what is Lagranges identity?

Answer 3

Just expand it out. The cross-product terms cancel.

Answer 4

Assuming the equation is correct, it must be true since each side of the equation must equal one another.

Answer 5

Well first add the initiating computational personficying clakitoode and do the cha cha cha.

Answer 6

expand both the sides man

Answer 7

Unfactored:
(a^(m + n)) + ((a^m)*(y^2)) + ((b^2)*(a^n)) + ((b^2)*(y^2)) = (a^(m + n)) + ((b^2)*(y^2)) + ((y^2)*(a^m)) + ((a^n)*(b^2))

Simplifies to:
((b^2) + (a^m))*((a^n) + (y^2)) = ((b^2) + (a^m))*((a^n) + (y^2))

Answer 8

if you can answer that give me a hala

Answer 9

I wish I could...!

Answer 10

i don't now