By what distance, x2, should the second child compress the spring so that the marble lands in the middle of the box? (assume that height of the box is negligible, so that there is no chance that the marble will hit the side of the box before it lands in the bottom.)]

u get ma big blacc balls

hello from mrbilldoesmath!

answer:

(7 x + 6 y) (49 x^2 - 42 x y + 36 y^2)

discussion:

the trick is to recognize that 343 = 7^3 and 216 = 6^3 and, in general,

a^3 + b^3 =   (a + b) (a^2 - a b + b^2).

in our case

343x^3+216y^3   =   (7x)^3 + (6y)^3

so take a = 7x and b = 6y in the above equation to get the answer.

you,

mrb

answer: none of these answers are correct only the second half is

step-by-step explanation: