Find a, b such that a^3 + ab^2 = 30, b^3 + ba^2 = 90

[Math_Junkie]

Here's a nice little brainteaser:

What is the only pair of real numbers which satisfies both:

a^3 + ab^2 = 30 and b^3 + ba^2 = 90

 

[stapel]

What have you tried? How far have you gotten?

Hint: Subtract the two equations, and factor, as best you can, the variable-containing side. Then try to solve one of the original equations for "(a² + b²) = (something)". Plug the (something) into the factored result of the subtraction, and see where that leads you. :wink:

Eliz.

 

[Math_Junkie]

Trust me, I've tried.
But I've come up with dead ends.

 

[skeeter]

a^3 + ab^2 = 30 and b^3 + ba^2 = 90

factor the left side of both equations ...

a(a^2 + b^2) = 30 and b(b^2 + a^2) = 90

note that both a and b cannot = 0, so divide all terms in one equation by the terms in the other equation ...

b(b^2 + a^2) = 90
----------------------
a(a^2 + b^2) = 30

b/a = 3/1

b = 3a

now ... proceed toward the unique solution for a and b.

 

[stapel]

Trust me, I've tried.
But I've come up with dead ends.

Please reply showing what you've done so far, and how far you got in following the "Hint" provided earlier.

Please be complete. Thank you!

Eliz.

 

[Math_Junkie]

Well, I got to where skeeter got and ended up with b = 3a.
I then plugged in the 3a where the b's were and got:

a^3 + ab^2 = 30
a^3 + a(3a)^2 = 30
a^3 + a(9a^2) = 30
a^3 + 9a^3 = 30
10a^3 = 30

and

b^3 + ba^2 = 90
(3a)^3 + (3a)a^2 = 90
27a^3 + 3a^3 = 90
30a^3 = 90

I'm not quite sure these are the right steps to take.
This was a question on a math contest from last year and I wanted to try and figure out by myself. :shock:

 

[skeeter]

if 10a^3 = 30, then a^3 = 3 ... a is the cube root of 3.