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

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?

#" "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...