(x+5)⁴+(x+3)⁴=16 how to solve it

y=x+4(1)

Then we have:

(y−1)4+(y+1)4=16

Expanding binomials, odd powers of y cancel, giving

2y4+12y2+2=16

⟹2y4+12y2−14=0

solve this quadratic in y2 for y2 then take square roots on both sides:

y=±−12±122+4⋅2⋅14−−−−−−−−−−−√4−−−−−−−−−−−−−−−−−−−√

substitute into (1):

x=−4±−12±122+4⋅2⋅14−−−−−−−−−−−√4−−−−−−−−−−−−−−−−−−−√

If you evaluate this for all four ± combinations the two real roots turn out to be −5 and −3.

Step-by-step explanation:

$15 each

step-by-step explanation:

$45/3 children= $15 each