Hectic Algbraic Eqns: a^3 + b^3 = 91, a^2 + b^2 = 61

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gautam

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Can any one help to find the solution of the problem

a[sup:2gpavzjg]3[/sup:2gpavzjg]+b[sup:2gpavzjg]3[/sup:2gpavzjg]=91
a[sup:2gpavzjg]2[/sup:2gpavzjg]+b[sup:2gpavzjg]2[/sup:2gpavzjg]=61

I am really bothered two find the method of this problem the answer will be 6 and -5 of correspoind value of "a" and "b" and vice-versa. But if any one can help to mention the method of this problem

Thanks Waiting for positive response
 
gautam said:
Can any one help to find the solution of the problem

a[sup:3g4flbbw]3[/sup:3g4flbbw]+b[sup:3g4flbbw]3[/sup:3g4flbbw]=91
a[sup:3g4flbbw]2[/sup:3g4flbbw]+b[sup:3g4flbbw]2[/sup:3g4flbbw]=61

I am really bothered two find the method of this problem the answer will be 6 and -5 of correspoind value of "a" and "b" and vice-versa. But if any one can help to mention the method of this problem

Thanks Waiting for positive response
Click to expand...

Since you do not have any particular method to follow - easiest would be graphical solution.

graph

\(\displaystyle y \, = \, (91\, - \, x^3)^{\frac{1}{3}}\)

and

\(\displaystyle y \, = \, \pm\,(61\, - \, x^2)^{\frac{1}{2}}\)

then find the intersection points.
 
Hello, gautam!

I haven't found a satisfactory method yet . . .

\(\displaystyle \text{Solve: }\; \begin{array}{cccc}a^3 + b^3 &=& 91 & [1]\\ a^2 + b^2 &=& 61 & [2]\end{array}\)
Click to expand...

\(\displaystyle \text{Factor [1]: }\;(a+b)(a^2 - ab + b^2) \:=\:91\)

\(\displaystyle \text{We have: }\;(a+b)(\underbrace{a^2+b^2}_{\text{This is 61}}-ab) \:=\:91\)

. . . . . . . . . . .\(\displaystyle (a + b)(61 - ab) \:=\:91\)

Now what?


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Out of desperation, I made a wild assumption.

. . \(\displaystyle a + b \:=\:1 \quad\Rightarrow\quad b \:=\:1-a \;\;{\bf[1]}\)

\(\displaystyle \text{The equation becomes: }\;61 - ab \:=\:91\)

\(\displaystyle \text{Substitute {\bf[1]}: }\;61 - a(1-a) \:=\:91 \quad\Rightarrow\quad a^2-a-30\:=\:0\)

. . \(\displaystyle (a - 6)(a + 5) \:=\:0 \quad\Rightarrow\quad a \:=\:6,-5\)

\(\displaystyle \text{Therefore: }\;(a,b) \;=\;(6,-5)\,\text{ or }\,(-5,6)\)


Of course, the assumption is totally unjustified.
. . All this is nonsense . . . *sigh*

 
Can any one help to find the solution of the problem

a[sup:bamn10zp]3[/sup:bamn10zp]+b[sup:bamn10zp]3[/sup:bamn10zp]=91
a[sup:bamn10zp]2[/sup:bamn10zp]+b[sup:bamn10zp]2[/sup:bamn10zp]=61

I am really bothered two find the method of this problem the answer will be 6 and -5 of correspoind value of "a" and "b" and vice-versa. But if any one can help to mention the method of this problem

Since 61 ends in 1, it might be a square. Taking the square root clearly verifies this.
Since 61 ends in 1, it might be a prime.
Dividing by all primes less than sqrt(61) = 7.81 will verify whether it is prime.
The calculations verify that 61 = 61 is prime.
Being of the form 4n + 1, it is therefore expressable as the sum of 2 squares (x^2 + y^2 = N)in one way only.
The minimum possible value of "x" is sqrt[61/2] = 5.52.
The maximum possible value of "x" is sqrt[4177] = 7.81.
Therefore, 5 <= x =< 7.

One quick look tells us that a and b are +/-5 and +/-6 giving us 5^2 + 6^2 = 61. We must now derive the proper signs of a and b to satisfy a^3 + b^3 = 91..

A couple of quick calculations tells us that a = -5 and b = +6 (or visa versa).yielding (-5)^3 + (6^)^3 = -125 + 216 = 91. =
 
following up on Soroban's response:

If 'a' and 'b' are integers then:

(a+b)(61-ab) = 13 * 7 = 1*91

we have following choices:

a+b = -1 and ab = 152 <<<<<<<<< no solution

a+b = 1 and ab = -30<<<<<<<<a = -5 b= 6 and a = 6 and b = -5

a+b = -91 and ab = 62

and so on...
 
Well, since a^2 + b^2 = 61, and we assume integers, then a^2 = 1 or 4 or 9 or 16 or 25 or 36 or 49 ...whoa! 64 too high!

Only possible combo is 25 (+-5) and 36 (+-6).
 
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