Question

If `a^3+b^3=8` and `a^2+b^2=4` what is the value of `(a+b)`?

Answer 1

There are two possible values for the sum, `a+b=2` (for `a=2` and `b=0`) or `a+b=-4` (for `a=-2 + i sqrt{2},` `b=-2 - i sqrt{2}).`

Explanation:

There are really two unknowns, the sum and the product of `a` and `b,` so let `x = a+b` and `y = ab`.

`x^2 = (a+b)^2 = a^2 + 2ab + b^2 = 2y + 4`

`x^3 = (a+b)^3 = a^3 + b^3 + 3ab(a+b) = 8+3 xy`

Two equations in two unknowns,

`2y = x^2 -4`

`2x^3 = 16 + 3x(2y) = 16 + 3x(x^2 - 4)`

`x^3 -12 x + 16 = 0`

That's called a depressed cubic, and those have a pretty easy closed form solution like the quadratic formula. But rather than touch that, let's just guess a root by the time honored method of trying small numbers. We see `x=2` works so `(x-2)` is a factor.

`x^3 -12 x + 16 = (x-2)(x^2 - 2x + 8) = 0`

We can now further factor

`x^3 -12 x + 16 = (x-2)(x-2)(x+4) = (x-2)^2(x+4) = 0`

So there are two possible values for the sum, `a+b=2` and `a+b=-4.`

The first answer corresponds to the real solution `a=2, b=0` and by symmetry `a=0, b=2`. The second answer corresponds to the sum of a pair of complex conjugates. They're `a,b=-2 \pm i sqrt{2}`. Can you check this solution?

Answer 2

`(a+b)=2, or, a+b=-4`

Explanation:

`" "a^2+b^2=4`

`=>(a+b)^2-2ab=4`

`=>2ab=(a+b)^2-4`

`=>ab=((a+b)^2-4)/2`

Now,

`" "a^3+b^3=8`

`=>(a+b)(a^2-ab+b^2)=8`

`=>(a+b)(4-ab)=8`

`=>(a+b){4-((a+b)^2-4)/2}=8`

`=>(a+b){6-((a+b)^2)/2}=8`

Let,

`(a+b)=x`

So,

`=>x(6-x^2/2)=8`

`=>x(12-x^2)=16`

`=>x^3-12x+16=0`

Observe that `2^3-12*2+16=8-24+16=0`

`:. (x-2)` is a factor.

Now, `x^3-12x+16=ul(x^3-2x^2)+ul(2x^2-4x)-ul(8x+16)`,

`=x^2(x-2)+2x(x-2)-8(x-2)`,

`=(x-2)(x^2+2x-8)`,

`=(x-2)(x+4)(x-2)`.

`:.x^3-12x+16==0 rArr x=2, or, x=-4`.

`:. a+b=2, or, a+b=-4`.

Graph is given here.

The value of `color(red)((a+b)=2, or, -4.`

Hope it helps...
Thank you...