## dtan5457 one year ago Isaac has written down three integers. The sum of all 3 when cubed is 7624. The sum of the range of the numbers (smallest and largest) is 27. The middle integer is 13. What are the other integers?

1. welshfella

(x + y + z)^3 = 7624 y = 13 x + z = 27

2. dtan5457

i have that im trying to solve x^3+z^3=7455 x+y=27

3. dtan5457

*x+z=27

4. welshfella

u sure you have the question right? 7624 is not a perfect cube

5. dtan5457

i think they mean the sum of each cube

6. mathstudent55

@welshfella I think you need to do x^3 + y^3 + z^3 = 7624

7. welshfella

x^3 + y^3 + z^3 = 7624

8. welshfella

ok

9. dtan5457

yea

10. anonymous

Isaac has written down three integers. The sum of all 3 when cubed is 7624. The sum of the range of the numbers (smallest and largest) is 27. The middle integer is 13. What are the other integers? WLOG assume $$a\le b <c$$ $$a^3+b^3+c^3=7624\\a+c=27\\b=13\\\implies a+b+c=40$$ now observe $$(a+b+c)^3=a^3+b^3+c^3+3(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2)+6abc\\40^3=7624+3(13a^2+169a+a^2c+ac^2+169c+13c^2)+78ac$$

11. welshfella

as a first step I would try x = 12 and z = 15 and x = 11 and z = 16

12. welshfella

trial and error

13. dtan5457

is it possible to solve algebraically?

14. mathstudent55

x^3 + 13^3 + z^3 = 7624 x^3 + 2197 + z^3 = 7624 x^3 + z^3 = 5427 x + z = 27

15. welshfella

weel you can generate an equation in x by plugging in x = 27 - z

16. anonymous

now since $$a+c=27$$ we get $$(a+c)^2=a^2+c^2+2ac\implies a^2+c^2=27^2-2ac$$ so consider now $$13a^2+169a+169c+13c^2+a^2c+ac^2+78ac\\\quad =169(a+c)+13(a^2+c^2)+ac(a+c+78)$$

17. welshfella

sorry z = 27 - x

18. welshfella

x^3 + (27-x)^3 = 7455

19. anonymous

oops, that should be $$3(13a^2+169a+169c+13c^2+a^2c+ac^2)+78ac$$ so $$3\cdot (169(a+c)+13(a^2+c^2)+ac(a+c+26))$$

20. anonymous

we have that $$a+c=27\implies a+c+26=53$$ and further that $$a^2+c^2=729-2ac$$ so $$64000=7624+3\cdot(169\cdot27+13\cdot(729-2ac)+ac\cdot53)$$

21. dtan5457

Never mind guys, i solved this using some quadratic formula. thanks for your help though

22. anonymous

so $$\frac{64000-7624}3-169\cdot27-13\cdot729=27ac\\\implies ac=176$$

23. dtan5457

I just did x^3+13^3+z^3=7624 x^3+z^3=5427 x+z=27 x=(-z+27) (-z+27)^3+z^3=5427 -z^3+81z^2-2187z+27^3+z^3=5427 81z^2-2187z+14256=0 ^^quadratic formula and you get the other 2 integers

24. mathstudent55

x = 27 - z x^3 = (27 - z)^3 x^3 = -z^3 + 81z^2 - 2187z + 19683 -z^3 + 81z^2 - 2187z + 19683 + z^3 = 5427 81z^2 - 2187z + 19683 = 5427 81z^2 - 2187z + 14256 = 0 z^2 - 27z + 176 = 0 (z - 11)(z - 16) = 0 z = 11 or z = 16

25. welshfella

i8 tried that and got -27x^2 as first term - must have made a mistake somewhere

26. welshfella

yea i see where i went wrong

27. anonymous

now we know $$ac=176\\a+c=27$$ so$$a(27-a)=176\\27a-a^2=176\\a^2-27a+176=0\\a=\frac{27\pm\sqrt{729-2\cdot176}}2\\\implies a=11\\\implies c=16$$

28. anonymous

but yeah, you can plug in for $$b$$ at the start and get $$a^3+c^3=\dots\\a^3+c^3=(a+c)^3-3ac(a+c)$$ and get $$ac$$ much faster

29. anonymous

thing is, once you have $$a+c,ac$$, writing out the quadratic and factoring is unnecessary because you ask the same question -- what two numbers add up to $$27$$ but multiply to give $$176$$?

30. dtan5457

you are right but i guess its just a little easier to go with things we learned before sometimes

31. dtan5457

even if it takes longer