Help me with this hard math problem please to study for math team!?

Question

The sum of two numbers is 29, and the cube of their sum exceeds the sum of their cubes by 14616. Find the larger of the two numbers.

Answer 1

so 2 equations
x + y = 29
(x + y)^3 = x^3 + y^3 + 14616

simplify the second one

x^3 + 3x^2y + 3xy^2 + y^3 = x^3 + y^3 + 14616
3x^2y + 3xy^2 = 14616
x^2y + xy^2 = 4872
xy( x+ y) = 4872
xy(29) = 4872
xy = 168
x(29 - x) = 168
i skipped some steps and you get
0 = x^2 - 29x +168
factor (x-21) (x-8)
answers 21, 8

Answer 2

So:

x+y = 29

(x+y)^3 = x^3 +y^3 + 14616.

Get out your calcuator:

(x+y)^3 = x^3+3xy^2 + 3x^2y + y^3

So replacing the expansion:

x^3+3x^2y+3xy^2+y^3 = x^3+y^3 + 14616

Subtract out common terms

3x^2y + 3xy^2 = 14616

now y = 29 -x, so:

3x^2(29-x) + 3x(29-x)^2 = 14616. Oh, we can divide by three and get slightly smaller numbers:

Let's do that, and expand the terms again:

29x^2 -x^3 + x(781-58x +x^2) = 4872

Combine terms:

-29x^2 +781x = 4872

Let's divide by 29 for sanity's sake

-x^2 +29x = 168

Multiply by -1 on both sides and collect all terms to the left

x^2 -29x +168 = 0

We can now factor

(x-21)(x-8)

Numbers are 8, 21.

Answer 3

Hey,

The two numbers are 21 and 8. The way I solved it is by first subtracting the given by the cubed root of 29, then I worked the cubed root of 9773 and found it through trial and error. Hope this helps, good luck.

Answer 4

I got a bad feeling about this . . .
x + y = 29
x^3 + y^3 + 14,616 = 24,389
x^3 + y^3 = 9,773
x^3 + (29 - y)^3 + 14,616 = 24,389
x^3 + 24,389 - 2,523x + 87x^2 - x^3 + 14,616 = 24,389
87x^2 - 2,523x + 14,616 = 0
x^2 - 29x + 168 = 0
168 = 2*2*2*3*7
(x - 8)(x - 21) = 0
x = 8, 21
y = 21, 8

x^3 + y^3 = 9,773
512 + 9261 = 9773
Either way, the larger number is 21.

Answer 5

21

Answer 6

John T gives a nice solution. To be honest, though, in competition I think the best approach is probably trial and error - there aren't that many possible solutions, and you'll probably hit on the right one after a few tries. Plus, the calculations are easier than simplifying the messy equation.