Four whole numbers are each rounded to the nearest 10
The sum of the four rounded numbers is 90
What is the maximum possible sum of the original four numbers?

Step-by-step explanation:

90÷4=22.5

30+20+20+20=90

original number with greatest value are 34,24,23,22

greatest sum=34+24+23+22=103

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