Question

Prove that (a²+b²)(c²+d²) = (ac+bd)² + (a-bc)²

Answer 1

Step-by-step explanation:

How can you prove that ? ... You evaluate the given numerical expression, (2-2)², by using the correct order of operations. ... Given: a²+b²= 2 and c²+d²=1 ... =(a^2+b^2)(d^2+c^2) , proved .

Answer 2

Answer:

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• How can I prove that (a2+b2)(c2+d2)=(ad−bc)2+(ac+bd)2?
• You can use complex numbers to quickly solve this.

• Use i2=−1

• (a2+b2)(c2+d2)=[a2−(−b2)][c2−(−d2)]

• ⟹ (a2+b2)(c2+d2)=[a2−(ib)2][c2−(id)2]

• ⟹ (a2+b2)(c2+d2)=[(a−ib)(a+ib)][(c+id)(c−id]

• ⟹ (a2+b2)(c2+d2)=[(a+ib)(c−id)][(a−ib)(c+id]

• ⟹ (a2+b2)(c2+d2)=[ac+i(bc−ad)−i2bd)][ac−i(bc−ad)−i2bd]

• ⟹ (a2+b2)(c2+d2)=[(ac+bd)+i(bc−ad)][(ac+bd)+i(bc−ad)]

• ⟹ (a2+b2)(c2+d2)=[(ac+bd)2−i2(bc−ad)2]

• ⟹ (a2+b2)(c2+d2)=(ac+bd)2+(bc−ad)2

• ⟹ (a2+b2)(c2+d2)=(ad−bc)2+(ac+bd)2

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