**so we can get 2 equations here, please ask if u want to know how i got these.**

**\(4a + 4b = 16\)**

**\(a^2 + 2ab + b^2 = 16\)**

**so now we combine them**

**\(4a + 4b = a^2 + 2ab + b^2\)**

**\(4a + 4b - 2ab - 2b^2 = (a+b)(a-b)\)**

**more on how that works if you ask me.**

**\(4(a+b) - 2b(a+b) = (a+b)(a-b)\)**

**\(4-2b = a-b\)**

**i think thats correct..**

**\(a+b = 4\)**

**oh well.....................**

**thats depressing oofoof**

**okok for b) its just 4, literally**

**and for c i don't think any integers work therfore no real numbers work becuase the max number for a or b is 1, becuase 2^4 is already 16 and negatives^4 are always postive for i think c is none.**

**and for a) when i did the failed equations u see a 2ab when i squared the first equation,try to single that out and u will get the answer. for a**

**HINT. for a, do \(2ab = a^4+b^4-a^2 - b^2\)**

**that will be equal to \(2ab = (a^2)(a^2 - 1) + (b^2)(b^2 - 1)\)**

**then use the second equation to substiture the a^2 and b^2 and u will get somewhere...**