Question

The number of ordered pairs of real numbers (a b) for which 2sqrt(x+3y)+(2)/(sqrt(x+3y))=5 and x=ay+b​

Answer 1

$Solution$ :

Observe that

(a+2b−3c)+(b+2c−3a)+(c+2a−3b)=0.

Because

x+y+z=0

$implies$

${x}^{3} + {y}^{3} + {z}^{3 } = 3xyz$

$we$$obtain$

${a + 2b - 3c}^{3} + {b + 2c - 3a}^{3} + {c + 2a - 3b}^{3}$

=$3(a+2b−3c)(b+2c−3a)(c+2a−3b).$