Show that (a^2+b^2)(c^2+d^2)=(ac+bd)^2 + (ad-bc)^2
Foil Left side gives you:
= a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2
= (ac)^2 + (ad)^2 + (bc)^2 + (bd)^2
Foil Right side (ac+bd)^2
= ac x ac + ac x bd + ac x bd + bd x bc
= (ac)^2 + 1acbd + 1acbd + (bd)^2
Foil (ad-bc)^2
= ad x ad - bc x ad - bc x ad + bc x bc
= (ad)^2 - 1acbd - 1acbd + (bc)^2
Combine all terms on right side
left with (ac)^2 + (bd)^2 + (ad)^2 + (bc)^2
To prove the given identity, we can expand both sides of the equation and show that they are equal.
Starting with the left-hand side (LHS):
LHS = (a^2 + b^2)(c^2 + d^2)
Expanding this using the distributive property, we get:
LHS = a^2(c^2 + d^2) + b^2(c^2 + d^2)
Now, let's expand each term:
LHS = a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2
Now, let's move on to the right-hand side (RHS):
RHS = (ac + bd)^2 + (ad - bc)^2
Expanding this, we get:
RHS = (ac + bd)(ac + bd) + (ad - bc)(ad - bc)
Using the distributive property, we can simplify this further:
RHS = a^2c^2 + 2abcd + b^2d^2 + a^2d^2 - 2abcd + b^2c^2
Now, let's simplify the RHS by combining like terms:
RHS = a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2
After simplifying both the LHS and RHS, we see that they are equal. Hence, we have proven that (a^2 + b^2)(c^2 + d^2) = (ac + bd)^2 + (ad - bc)^2.
Therefore, the given identity is true.