the sum of the squares of complex numbers X and Y is 7 and their cube is 10,what is the largest value x+y have

Question

please show work

Answer 1

We know that (1): x^2 + y^2 = 7 and (2): x^3 + y^3 = 10. We can factor the second eqn as

(x+y)(x^2 - xy + y^2) = 10 and substitute to get
(3): (x+y)(7 - xy) = 10.

Now we need to express xy in terms of x+y. To do this we can manipulate eqn (1) to become

(x+y)^2 - 2xy = 7. Solving for xy gives
xy = ((x+y)^2 - 7)/2.

We can substitute this result into eqn (3). Our penultimate step is to let x+y = t, which transforms eqn (3) into a cubic eqn with "nice" roots. I'll leave the rest of the algebra to you. ;)

Note that we didn't need to solve for x and y explicitly which probably would have been messier.

Answer 2

x² + y² = 7 implies (x² + y²)³ = 343 and (x³ + y³)³ = one hundred, for this reason 243=(x² + y²)³-(x³ + y³)²=x²*y²*(21-2*x*y) to that end x²*y²*(21-2*x*y)-243=0. Factoring, we get (2*x*y-9)*(x*y-9)*(x*y+3)=0, for this reason x*y ought to equivalent the two 9/2,9 or -3. on the grounds that x² + y² = 7, x^4 + x²*y² = 7*x² so we ought to have the two x^4 + (9/2)² = 7*x², x^4 + 80 one = 7*x² or x^4 + 9 = 7*x². employing the quadratic formula it rather is trouble-free to locate the corresponding opportunities for the fee of x², and because x² + y² = 7, the corresponding values for y². the optimal fee of Re(x+y) is then straight forward to locate. If I certainly have extra time later i visit supply the respond explicitly.

Answer 3

x^2 + y^2 = 7
... (a+bi)^2 + (c+di)^2 = 7
... a^2 +2abi -b +c^2 +2cdi -d = 7
... a^2 +c^2 +2i(ab+cd) -b -d = 7

x^3 + y^3 = 10
(a+bi)^3 + (c+di)^3 = 10
(a^2+2abi+bi^2)(a+bi)= (a^3 +2a^2*bi -a +a^2*bi +2a*bi^2 -b^3i) + (a^3 +3a^2*bi -a -2a*b^2 -b^3i) + (c^3 +3c^2*di -c -2c*d^2 -d^3i)

... i seem to have become lost
i was going to suggest subtracting one from the other for the result (equaling 3) but i have gone astray methinks and i have to leave my desk.