Question

If a^3+b^3=35 ,a^2+b^2=13. Find the value of a and b.

Answer 1

An evident solution is a=2, b=3
and
a=3, b=2

Let us find another solutions.

From
(*) a² + b²=13we get
a² + b² + 2ab = 13 + 2abso
(%) (a+b)² = 13 + 2abMoreover, from

(**) a³ + b³ = 35we get

35 = a³ + b³ = (a+b) (a² + b² - ab) = and using (*) we obtain(%%) 35 = (a+b) (13-ab).

Make a substitution:

s = ab
t = a+b.

Then we get from (%) and (%%):

t² = 13+2st(13-s)=35

Then
s = (t²-13)/2 whence
t (13 - (t² - 13)/2 )=35By elementary transformations this reduces to the equation:

(***) t³ - 39t + 70 = 0.Since we know one solution a=3 and b=2, it is easy to verify that
t=a+b=5 is a root of (***), whence this equation is divided by t-5.

Dividing we obtain that
t³ - 39t + 70 = (t-5)(t²+5t-14) = 0Solving
t²+5t-14=0we get
t=2 and t=-7.

Hence we have three roots: -7, 2, 5.

1) Let t=-7.
Then s = (t²-13)/2 = (49-13)/2 = 18, so s = ab = 18
t = a+b = -7

Substitutin a = -7 - b into the first equation we get
(-7 - b)b = 18
b² + 7b + 18 = 0D = 4*49 - 18² = -128 < 0, so there are no solutions.


2) Let t = 2.
Then s = (t²-13)/2 = (4-13)/2 = -9/2 = -4.5, so s = ab = -4.5
t = a+b = 2

Substitutin a = 2 - b into the first equation we get
(2-b)b = -4.5
b² - 2b - 4.5 = 0D = 4 + 4*4.5 = 22
whence

b₁ = (2 + √(22))/2b₂ = (2 - √(22))/2whence

a₁ = b₂ = (2 - √(22))/2a₂ = b₁ = (2 + √(22))/23) If t=5, then
s = (t²-13)/2 = (25-13)/2 = 6, so s = ab = 6
t = a + b = 5

Then similar calculations shows that
a = 3 and b = 2
or
a = 2 or b = 3.

Thus we have four solutions:

(2,3),
(3,2),
( (2 - √(22))/2, (2 + √(22))/2 ),
( (2 + √(22))/2, (2 - √(22))/2 ).