Question

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

Answer 1

to solve these kind of questions, all we have to do is choose one side of the equation and play around with it till it becomes the other side. In this case I've already tried starting with the left hand side but it's difficult. So we'll start with the right hand side:

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

Answer 2

let's expand it out (because you just have to try something to make it equal to (a^2+b^2)(c^2+d^2)

we get \[a ^{2}b^{2}+2abcd+b^{2}d^{2}+a^{2}d^{2}-2abcd+b^{2}c^{2}\]

Answer 3

which equals to \[a^{2}b^{2}+b^{2}d^{2}+a^{2}d^{2}+b^{2}c^{2}\]

so from here we factorise any common factors we can find. so over here i notice that \[a^{2}b^{2} \] and \[a^{2}d^{2}\] share the common factor \[a^{2}\]. so I'll factorise a^2 out to get \[a^{2}(c^{2}+d^{2})+b^{2}d^{2}+b^{2}c^{2}\]

Answer 4

and similarly we factorise out b^2 to get \[a^{2}(c^{2}+d^{2})+b^{2}(d^{2}+c^{2})\]
and finally we factorise out the (c^2+d^2) to get (a^2+b^2)(c^2+d^2)

SOLVED!

Answer 5

Thank you soooo much!!