Question

a^3 - 3a^2 + 5a = 1
b^3 - 3b^2 + 5b = 5
a + b = ?

Answer 1

\[a^3-b^3-3a^2+3b^2+5a-5b=1-5 \\ (a-b)(a^2+ab+b^2)-3(a-b)(a+b)+5(a-b)=-4 \\ (a-b)[(a^2+ab+b^2)-3(a+b)+5]=-4 \\ (a-b)(a^2+ab+b^2-3a-3b+5)=-4 \\ \]
So far this is what I thought up..
I'm not sure if I will continue on this path or change it up.

Answer 2

I went the other way and aded them up but did not get me anywhere.

Answer 3

and I guess we are looking for a+b not a-b anyways

Answer 4

Each has one real root and two complex roots. If you add up the corresponding roots of each equation they all add up to 2!

Answer 5

I know the three roots of a cubic equation, ax^3 + bx^2 + cx + d, add to -b/a.
Their product is -d/a.

Answer 6

if it is so, the a from your last equation is 1 , b = -3, c=5 and d =-1, the original one is a^3 -3a^2 +5a -1 =0 have 2 roots a1, a2 which satisfy a1+a2= 3 and a1*a2= 1

Answer 7

Do I mess it up? You use the same letter to indicate to 2 different problems. hehehe... that makes me dizzy

Answer 8

Cubic equation have three roots. So a1 + a2 + a3 = 3; a1*a2*a3 = 1

Answer 9

anyway, your cubic equation don't have integer solutions. I don't know how to find them out but (a + b) = 1/ the rest of a equation + 5/ the rest of b equation
I factor a , b out , isolate them, then add them up. hehehe... stupid way but I don't know how to do more.

Answer 10

This was an exam question where no calculators or graphing devices allowed. So they don't expect people to find the roots to answer this question.

Answer 11

If the two equations can be combined and expressed as follows, then we may be able to solve it: (a+b)^3 + k1(a+b)^2 + k2(a+b) + k3 = 0 then this will have one rational root, namely, a+b = 2.

Answer 12

too complicated. For which grade the problem is ?

Answer 13

This was for a college entrance exam in India that my nephew took last year.

Answer 14

Wow!! India is the leader on Math and computer.

Answer 15

This particular engineering college is tough to get into and each year they come up with questions in Math, Physics, Inorganic and Organic Chemistry that is just awesome.

Answer 16

Agree!! However, awesome if we can solve it. If we cannot, no more awesome, hehehe.. I think so.

Answer 17

Oh, this will get solved eventually. I will just forget about it and go about my business and sooner or later the answer is likely to pop up. Or someone here may be able answer this.

Answer 18

any restrictions on a, b ? are they integers ?

Answer 19

Ok so let's see...have you tried solving for 'a' and 'b' simultaneously?

Answer 20

If a1, a2, a3 are the three roots of a^3 - 3a^2 + 5a - 1 = 0, then:
a1 + a2 + a3 = 3
a1 * a2 * a3 = 1
a1*a2 + a2*a3 + a3*a1 = 5

If b1, b2, b3 are the three roots of b^3 - 3b^2 + 5b - 5 = 0, then:
b1 + b2 + b3 = 3
b1 * b2 * b3 = 5
b1*b2 + b2*b3 + b3*b1 = 5

Answer 21

We want a1+b1 or a2 + b2 or a3 + b3.

Answer 22

lol i am back at the adding step...

Answer 23

No restrictions given in the problem for a and b. But we cannot assume they are integers. Each equation has one real, non-integer root and two complex roots if I use a graphing tool. But it is not allowed.

Answer 24

if it helps,
answer is a+b = 2 :

http://www.wolframalpha.com/input/?i=max+and+min+a+%2B+b+%2C+a%5E3%2Bb%5E3-3%28a%5E2%2Bb%5E2%29%2B5%28a%2Bb%29%3D6

Answer 25

may be we can reverse engineer - show that the upper and lower bounds of a+b is 2

Answer 26

http://www.wolframalpha.com/input/?i=%28a%2Bb%29%5E3-3%28a%2Bb%29%5E2%2B5%28a%2Bb%29-3ab%28a%2Bb%29%2B6%28ab-1%29
there is also this

Answer 27

if we can some factor equ1+equ2 to into (a+b-2)(times whatever)=0
then we got it

Answer 28

Roots of the first equation are: 0.22908, 1.3855 + 1.5639i, 1.3855 - 1.5639i
Roots of the first equation are: 1.77092, 0.6145 - 1.5639i, 0.6145 + 1.5639i

Look how nicely each of the corresponding root pairs add to 2!

Answer 29

*second*

Answer 30

somehow*

Answer 31

but i guess that is what ranga said above somewhere anyways

Answer 32

wolf implicit plot gives a line hmmm

Answer 33

http://www.wolframalpha.com/input/?i=+a^3%2Bb^3-3%28a^2%2Bb^2%29%2B5%28a%2Bb%29%3D6

Answer 34

Oh my!

Answer 35

no idea why
i don't thing you can write this as a polynomial in \(a+b\) because \((a+b)^3\) and terms in it like \(ab^2\) that you can't get rid of with \((a+b)^2\) or \(a+b\) so i although it looks like the right approach, i have no idea how you can carry out that plan

Answer 36

Like I get this so far
\[(a+b)^3-3(a+b)^2+5(a+b)=3ab(a+b-2)+6\]

Answer 37

which is neat because the other thingys we had were in this form
you know
\[a^3-3a^2+5a=1 \text{ and } b^3-3b^2+5b=5\]

Answer 38

so we have f(a)=1 and f(b)=5
and f(a+b)=3ab(a+b-2)+6

Answer 39

max/min of \(f : a + b \)
subject to \(g : a^3+b^3-3(a^2+b^2)+5(a+b)=6\)

\(\nabla f = \lambda \nabla g \)
\(\langle1, 1 \rangle = \lambda \langle 3a^2 -6a + 5, 3b^2-6b+5 \rangle\)

\(\implies 3a^2 - 6a + 5 = 3b^2 - 6b + 5\)
\(\implies 3(a^2-b^2) - 6(a-b) = 0\)
\(\implies a=b ~ \lor ~~a+b = 2\)

Answer 40

^lagrange may be luxury for college entrance problem.. but thats all i can think of as of now..

Answer 41

wish i had thought of that

Answer 42

\[
(1):~~~a^3 - 3a^2 + 5a = 1 \\
(2):~~~b^3 - 3b^2 + 5b = 5 \\
\text{ }\\
(1):~~~a^3 - 3a^2 + 5a = 1 \\
\text{The }a^3 - 3a^2 \text{ terms suggest } (a-1)^3.~~~~~
(a-1)^3 = a^3 - 3a^2 + 3a - 1 \\
(1):~~~(a-1)^3 + 2a + 1 = 1 \\
(1):~~~(a-1)^3 + 2(a-1) + 2 + 1 = 1 \\
(1):~~~(a-1)^3 + 2(a-1) = -2 \\\\
\text{ }\\
(2):~~~b^3 - 3b^2 + 5b = 5. ~~~~
(b-1)^3 = b^3 - 3b^2 + 3b - 1 \\
(2):~~~(b-1)^3 + 2(b-1) + 3 = 5 \\
(2):~~~(b-1)^3 + 2(b-1) = 2 \\\\
\text{ }\\
\text{Let } f(x) = x^3 + 2x \\
\text{ }\\
(1): f(a-1) = -2 \\
(2): f(b-1) = 2 \\
\text{ }\\
f(x) \text{ is an odd function. That is, } f(x) = -f(-x) \\
\text{ }\\
\text{Therefore, from (1) and (2) above, } a - 1 = -(b - 1) \\
a-1 = -b + 1 \\
a+b = 2 \\
\]

Answer 43

beautiful !! xD

Answer 44

omg, ranga that is the "cutest" thing I ever seen

Answer 45

Cuter than kittens or puppies?! lol. Thank you all.

Here is another interesting problem from the same test. This has been solved and so I will close this thread.
\[\Large \text{If }~a, b, c, d ~ \in ~ I^+ \text{ and }~ \log_{a} b = \frac 32, ~ \log_{c} d = \frac 54 \text{ and } \\ \Large (a - c) = 9 \text{ then the remainder when }~b^d~\\ \text{ }\\\Large \text{is divided by 31 is _______.}\]

Answer 46

a,b,c,d are positive integers.

Answer 47

The best way I came up with his trial and error
and finally by fermats little theorem I got the answer is 1.

Answer 48

\[
b = a^{3/2};~~~ d = c^{5/4} \\
\text{ }\\
\text{If a,b,c,d are all integers, then 'a' has to be a perfect square and 'c' has to be } \\
\text{a perfect 4th power because we have to take the square root of 'a' and the }\\
\text{fourth root of 'c' to find b and d.}\\
\text{ }\\
\text{Let }a = p^2; ~~c = q^4 \text{ where p and q are integers.}\\
a - c = p^2 - q^4 = (p - q^2)(p + q^2) = 9 = 1 \times 9 \\
p - q^2 = 1; ~~~ p + q^2 = 9 \\
2p = 10; ~~~ p = 5; ~~~ q = 2; ~~~ a = 25; ~~~ c = 16; ~~~ b = 125; ~~~ d = 32 \\
\text{ }\\
b^d~\text{mod}~31 = 125^{32}~\text{mod}~31 = (125~\text{mod}~31)~\text{mod}~31 = 1~\text{mod}~31 = 1.
\]

Answer 49

Yes, freckles, you got the same final answer as mine!

Answer 50

In the last line an exponent was left out:
\[
b^d~\text{mod}~31 = 125^{32}~\text{mod}~31 = (125~\text{mod}~31)^{32}~\text{mod}~31 = 1~\text{mod}~31 = 1.
\]

Answer 51

\(a^{30}\equiv 1 \mod 31\)

Answer 52

interestin ! did u show \(30 | \frac{3}{2} c^{\frac{5}{4}} \) @freckles ?

Answer 53

wow ! @ranga 's solution looks very neat xD

Answer 54

Thank you @rational. Enjoy sharing these atypical problems.