Answer
$(a^{2}+b^{2})(c^{2}+d^{2})=(ac+bd)^{2}+(ad-bc)^{2}$
Work Step by Step
$(a^{2}+b^{2})(c^{2}+d^{2})=(ac+bd)^{2}+(ad-bc)^{2}$
Evaluate the powers on the right side:
$(a^{2}+b^{2})(c^{2}+d^{2})=a^{2}c^{2}+2acbd+b^{2}d^{2}+a^{2}d^{2}-2adbc+b^{2}c^{2}$
Cancel $2acbd$ and $-2adbc$ on the right side:
$(a^{2}+b^{2})(c^{2}+d^{2})=a^{2}c^{2}+b^{2}d^{2}+a^{2}d^{2}+b^{2}c^{2}$
Factor the right side by grouping terms and the identity will be proved:
$(a^{2}+b^{2})(c^{2}+d^{2})=(a^{2}c^{2}+a^{2}d^{2})+(b^{2}c^{2}+b^{2}d^{2})$
$(a^{2}+b^{2})(c^{2}+d^{2})=a^{2}(c^{2}+d^{2})+b^{2}(c^{2}+d^{2})$
$(a^{2}+b^{2})(c^{2}+d^{2})=(a^{2}+b^{2})(c^{2}+d^{2})$
