Question

If a+b+c = 0, where a,b,c are non zero real numbers, find the value of,

(a² - bc)² - (b²-ca) - (c²-ab)

Answer 1

 1) Since c = -(a+b), 

a²/(bc) + b²/(ca) + c²/(ab) 
= -a²/(b(a+b)) - b²/(a(a+b)) + (a+b)²/(ab) 
= [-a³ - b³ + (a+b)³] / (ab (a+b)) 
= (3a²b + 3ab²) / (ab (a+b)) 
= 3ab(a + b) / (ab (a+b)) 
= 3. 
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2) Note that (x² + bx + c)² = x⁴ + (2b) x³ + (b² + 2c) x² + (2bc)x + c². 

Comparing coefficients with x⁴ + 6x³ + 19x² + 30x + 25, 
2b = 6 ==> b = 3. 
b² + 2c = 9 + 2c = 19 ==> c = 5. 
(Check: 2bc = 30). 

So, we need to add c² = 5² = 25 to make a perfect square, and 
x⁴ + 6x³ + 19x² + 30x + 25 = (x² + 3x + 5)².