if the sum of the numbers is 8 and the sum of their cubes is 20 whats is the sum of their squares?

Question

Please help me its due tomorrow and it took me the whole day and i couldnot solve this question,,,,,,Thank u ..

Answer 1

Ok

you need to know that (a+b)^3 = a^3 + b^3 + 3ab(a+b), you can see that by expanding it out yourself.

So 8^3 = 20 + 3ab(8), you get ab = 41/2.

now look at (a+b)^2 = a^2 + b^2 + 2ab.

8^2 = a^2 + b^2 + 2(41/2)

You get a^2 + b^2 = 23.

Answer 2

you have that
x + y = 8
so you can substitute for y in the expression for the sum of cubes:
x^3 + y^3 = x^3 - (8-x)^3
this simplifies to
24x^2 - 192x + 512
it can be shown that this quadratic expression has a minimum at x = 4, where the sum of cubes is 128 (greater than 20!). so there is no real solution - are complex solutions allowed? the sum of squares may still be real, i haven't worked that out.

Answer 3

a + b = 8
a^3 + b^3 = 20
(8 - b)^3 + b^3 = 20. . . . . . . . answer for b is not an integer

Answer 4

a + b = 8
(a+b)² = 8²
a²+2ab+b² = 64

a³ + b³ = 20
(a+b)(a²-ab+b²) = 20
8(a²-ab+b²) = 20
(a²-ab+b²) = 5/2

a²+2ab+b² = 64
a²-ab+b² = 2.5
---------------------
3ab = 61.5
ab = 20.5

a²+b² = 2.5 + 20.5 = 23