Show that (a^2+b^2)(c^2+d^2)=(ac+bd)^2 + (ad-bc)^2
To prove the equation (a^2 + b^2)(c^2 + d^2) = (ac + bd)^2 + (ad - bc)^2, we can expand both sides of the equation and simplify.
Let's start with the left-hand side (LHS):
(a^2 + b^2)(c^2 + d^2)
= a^2(c^2 + d^2) + b^2(c^2 + d^2) (Using distributive property)
= a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2 (Expanding each term)
Now let's move on to the right-hand side (RHS):
(ac + bd)^2 + (ad - bc)^2
= (ac + bd)(ac + bd) + (ad - bc)(ad - bc) (Using the property (a + b)^2 = a^2 + 2ab + b^2)
= (ac)^2 + (ac)(bd) + (bd)(ac) + (bd)^2 + (ad)^2 - (ad)(bc) - (bc)(ad) + (bc)^2 (Expanding each term)
By rearranging the terms, we can simplify the RHS:
(ac)^2 + (ac)(bd) + (bd)(ac) + (bd)^2 + (ad)^2 - (ad)(bc) - (bc)(ad) + (bc)^2
= a^2c^2 + 2(ac)(bd) + b^2d^2 + a^2d^2 - 2(ad)(bc) + b^2c^2 (Rearranging terms)
Comparing the LHS and RHS, we see that they are the same:
LHS = a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2
RHS = a^2c^2 + 2(ac)(bd) + b^2d^2 + a^2d^2 - 2(ad)(bc) + b^2c^2
Since the LHS is equal to the RHS, we have proven that (a^2 + b^2)(c^2 + d^2) = (ac + bd)^2 + (ad - bc)^2.