# Prove this identity:(a^2 + b^2) (c^2 + d^2) = (ac +\- bd)^2 + (ad -\+ bc)^2

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Top Answer

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

(a^2 + b^2) (c^2 + d^2) = (ac +- bd)^2 + (ad -+ bc)^2

Let us start from the right side:

(ac + bd)^2 + (ad - bc)^2 = (ac)^2 + 2acbd + (bd)^2 + (ad)^2 - 2abcd + (bc)^2

Let us simplify:

(ax+bd)^2 + (ad-bc)^2 = (ac)^2 + (bd)^2 + (ad)^2 + (bc)^2

Let us rearrange terms:

==> (ac+ bd)^2 + (ad-bc)^2= (ac)^2 (bc)^2 + (bd)^2 + (ad)^2

Now we will factor:

= c^2 (a^2+b^2) + d^2(a^2+b^2)

= (a^2+ b^2)(c^2 + d^2)...

==> (ac+bd)^2 + (ad-ac)^2 = (a^2 + b^2 )(c^2 + d^2)

Similarly:

(ac-bd)^2 + (ad+ac)^2 = (a^2 + b^2)((c^2 + d^2)

neela | High School Teacher | (Level 3) Valedictorian

Posted on

To prove the identity:

(a^2 + b^2) (c^2 + d^2) = (ac +- bd)^2 + (ad -+ bc)^2.

Proof:

We take the right  side first and expand both terms:

(ac+bd)^2 = a^c^2+2abcd +b^2d^2..........(1)

We know that (ad-bc)^2 = a^2d^2-2abcd+b^2c^2^2....(2)

Adding the  both eq(1) and (2), we see that middle term 2abcd get cancelled on the right and  we get:

(ac+bd)^2+(ad-bc)^2 = a^c^2 +b^2d^2+a^2d^2+b^2c^2 (ac+bd)^2+(ad-bc)^2 = a^2c^2+a^2d^2+b^2d^2+b^2c^2

(ac+bd)^2+(ad-bc)^2 = a^2(c^2+d^2)+b^2(d^2+c^2)

(ac+bd)^2+(ad-bc)^2 = (c^2+d^2)(a^2+b^2) = LHS.

Therefore ,

(a^2 + b^2) (c^2 + d^2) = (ac + bd)^2 + (ad + bc)^2.

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