# Answer to Question #3646 in Algebra for Vikas Soni

Question #3646

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

Expert's answer

An evident solution is a=2, b=3

and

a=3, b=2

Let us find another solutions.

From

(*) a

a

(%) (a+b)

(**) a

35 = a

Make a substitution:

s = ab

t = a+b.

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

t

Then

s = (t

t (13 - (t

(***) t

t=a+b=5 is a root of (***), whence this equation is divided by t-5.

Dividing we obtain that

t

t

t=2 and t=-7.

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

1) Let t=-7.

Then s = (t

t = a+b = -7

Substitutin a = -7 - b into the first equation we get

(-7 - b)b = 18

b

2) Let t = 2.

Then s = (t

t = a+b = 2

Substitutin a = 2 - b into the first equation we get

(2-b)b = -4.5

b

whence

b

a

s = (t

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 ).

and

a=3, b=2

Let us find another solutions.

From

(*) a

^{2}+ b^{2}=13we geta

^{2}+ b^{2}+ 2ab = 13 + 2abso(%) (a+b)

^{2}= 13 + 2abMoreover, from(**) a

^{3}+ b^{3}= 35we get35 = a

^{3}+ b^{3}= (a+b) (a^{2}+ b^{2}- ab) = and using (*) we obtain(%%) 35 = (a+b) (13-ab).Make a substitution:

s = ab

t = a+b.

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

t

^{2}= 13+2st(13-s)=35Then

s = (t

^{2}-13)/2 whencet (13 - (t

^{2}- 13)/2 )=35By elementary transformations this reduces to the equation:(***) t

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

Dividing we obtain that

t

^{3}- 39t + 70 = (t-5)(t^{2}+5t-14) = 0Solvingt

^{2}+5t-14=0we gett=2 and t=-7.

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

1) Let t=-7.

Then s = (t

^{2}-13)/2 = (49-13)/2 = 18, so s = ab = 18t = a+b = -7

Substitutin a = -7 - b into the first equation we get

(-7 - b)b = 18

b

^{2}+ 7b + 18 = 0D = 4*49 - 18^{2}= -128 < 0, so there are no solutions.2) Let t = 2.

Then s = (t

^{2}-13)/2 = (4-13)/2 = -9/2 = -4.5, so s = ab = -4.5t = a+b = 2

Substitutin a = 2 - b into the first equation we get

(2-b)b = -4.5

b

^{2}- 2b - 4.5 = 0D = 4 + 4*4.5 = 22whence

b

_{1}= (2 + √(22))/2b_{2}= (2 - √(22))/2whencea

_{1}= b_{2}= (2 - √(22))/2a_{2}= b_{1}= (2 + √(22))/23) If t=5, thens = (t

^{2}-13)/2 = (25-13)/2 = 6, so s = ab = 6t = 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 ).

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