Question

solve the following system of equations
\[\large 1+a^3+3ab=b^3 \\ \large 1+a^5=b^5\] a,b are real numbers …

Answer 1

hint
\[\large x^3 + y^3 + z^3 - 3xyz = (x + y + z) (x^2 + y^2 + z^2 - xy -yz - zx)\]

Answer 2

From first by using this formula, I got:

b -a = 1

What to do next??

Answer 3

put it in second one

Answer 4

Second equation???
But how, I have no idea..

Answer 5

b=a+1
1+a^5=(a+1)^5

Answer 6

Not sure but it becomes:

\[(a+1)^5 - a^5 = 1\]

Is there any formula then??

Answer 7

Sorry it can also become as:

\[(a + 1)^5 = a^5 + 1^5\]

Answer 8

use binomial theorem to expand (a+1)^5

Answer 9

Oh I am also thinking the same..

Answer 10

\[(a+1)^5= a^{10} + 5a^4 + 10a^3 + 10a^2 +5a +32\]

Am I right??

Answer 11

1 not 32

Answer 12

and a^5 not a^10

Answer 13

Hint: There will be two values for a. (Is this correct?)

Answer 14

Oh sorry..
\[(a+1)^5= a^5 + 5a^4 + 10a^3 + 10a^2 +5a +1\]

Is this right now??

Answer 15

yes

Answer 16

So we have left with:

\[5a^4 + 10a^3 + 10a^2 +5a = 0\]

Answer 17

Now I have to make factors of this??

Answer 18

yep

Answer 19

I am having problem in making factors..
Can they form by grouping these terms or by putting 1. -1 etc etc??

Answer 20

(0,1) and (-1,0)

Answer 21

Yeah, I got one factor at a = -1..

Answer 22

grouping

Answer 23

u have
\[a^4+2a^3+2a^2+a=a^4+a+2a^3+2a^2\]

Answer 24

I have found it by Long Division Method..

Answer 25

\[(a+1)(5a^3 + 5a^2 + 5a) = 5a(a + 1)(a^2 + a + 1)\]

So, a= 0, a = -1 are the two solutions..

Finding other two also..

Answer 26

As the question says a and are real, so other two cannot be possible because they are complex..

\[a = \frac{-1 \pm i \sqrt{3}}{2}\]

Answer 27

@waterineyes
well done

Answer 28

So, a = 0 or it will be -1

According to it,

b = 1 or b = 0

So,

When a = 0, b = 1
when a = -1 b = 0 are the solutions..

Answer 29

No, @mukushla you guided me well...
Thanks for that really...

Answer 30

welcome my friend

Answer 31

Notice fermat's last theorem.
x^n + y^n = z^n have no solution(s) for n>2 and (x,y,z) are integers.
Applying to eq(2), one of a or b must be 0. and we'll get (0,1)(-1,0). Check eq(1) LHS will be same as RHS. so solutions are (0,1)(-1,0)

Is it possible?