Prove that (a^2+b^2)(A^2+B^2) = (aA+bB)^2 + (aB-bA)^2?

Question

Try to prove for all cases of a, b, A, B, where a, b, A, B are real numbers.

Answer 1

Simplify the left side:
a^2A^2 + a^2B^2 + b^2B^2 + b^2A^2
Simplify the right side:
a^2A^2 + 2abAB + b^2B^2 + a^2B^2 - 2abAB + b^2A^2
= a^2A^2 + b^2B^2 + a^2B^2 + b^2A^2
Left side = Right Side
Therefore the relation is true.

Answer 2

(a^2+b^2)(A^2+B^2) = (aA+bB)^2 + (aB-bA)^2
Multiply out the products:
a²A²+a²B²+b²A²+b²B²=a²a²+2aAbB+b²B²+a²B²-2aAbB+b²A²
Cancel the 2aAbB and rearrange:
a²A²+a²B²+b²A²+b²B²=a²A²+a²B²+b²A²+b²B² ← Tautology

Therefore (a^2+b^2)(A^2+B^2) = (aA+bB)^2 + (aB-bA)^2

Answer 3

(a^2+b^2)(A^2+B^2)
=(Aa)^2+(Ba)^2+(Ab)^2+(Bb)^2

(aA+bB)^2 + (aB-bA)^2
=(Aa)^2+(Bb)^2+2AaBb+(Ba)^2+(Ab)^2-2BaAb
=(Aa)^2+(Ba)^2+(Ab)^2+(Bb)^2
It is so easy. My 3 year old cousin could do it.

Answer 4

(a^2+b^2)(A^2+B^2)
a^2.A^2+a^2.B^2+b^2.A^2+b^2.B^2
note this step add and subtract 2aAbB
a^2.A^2+a^2.B^2+b^2.A^2+b^2.B^2+2aAbB-2aAbB
Rearranging
a^2.A^2+2aAbB+b^2.B^2+a^2.B^2+b^2.A^2-2aAbB
[a^2.A^2+2aAbB+b^2.B^2]+[a^2.B^2+b^2.A^2-2aAbB]
Hence
(aA+bB)^2+(aB-bA)^2

Answer 5

properly, no individual ever idea Oswald became conventional (I talked about a documentary on him) and do you imagine he did slightly reading and idea, "wow, faster or later... someone's' gonna ask a question about coincedence on Y!A".