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

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step-by-step explanation:

the correct answer is 828.07