Use the identity (x^2+y^2) ^2 = (x^2-y^2) ^2 + (2xy) ^2 to determine the sum of the squares of two numbers if the difference of the squares of the numbers is 5 and the product of the numbers is 6.

Step-by-step explanation:

(x^2+y^2) ^2 = (x^2) ^2+2x^2y^2 + (y^2) ^2

Adding and substracting 2x^2y^2

We get

(x^2+y^2) ^2 = (x^2) ^2+2x^2y^2 + (y^2) ^2 + 2x^2y^2-2x^2y^2

And we know a^2-2ab+b^2 = (a-b) ^2

So we identify (x^2) ^2 as a^2, (y^2) ^2 as b^2 and - 2x^2y^2 as - 2ab. So we can rewrite (x^2+y^2) ^2 = (x^2 - y^2) ^2 + 2x^2y^2 + 2x^2y^2 = (x^2 - y^2) ^2+4x^2y^2 = (x^2 - y^2) ^2+2^2x^2y^2

Moreever we know (a·b·c) ^2=a^2·b^2·c^2 than means 2^2x^2y^2 = (2x·y) ^2

And (x^2+y^2) ^2 = (x^2 - y^2) ^2 + (2x·y) ^2