Let x be the smallest of the 3 integers, then since the sum of the three consecutive even integers is 6356 we can set the following equation:
[tex]x^2+(x+2)^2+(x+4)^2=6356.^{}[/tex]
Simplifying the above equation we get:
[tex]\begin{gathered} x^2+x^2+4x+4+x^2+8x+16=6356, \\ 3x^2+12x+20=6356. \end{gathered}[/tex]
Then to answer this question we must solve the following equation:
[tex]3x^2+12x-6336=0.[/tex]
Using the quadratic formula for second-degree equations we get:
[tex]x=\frac{-12\pm\sqrt[]{12^2-4(3)(-6336)}}{2(3)}\text{.}[/tex]
Therefore:
[tex]x=44\text{ or x=}-48.[/tex]
Since the problem is asking for positive integers, then x=44, and the integers are 44, 45, and 46, therefore its sum is 135.
Answer: 135.
