Question

COMPLEX NUMBERS? - If (a+ib)^3 =8 , prove that a^2 + b^2 =4.

Answer 1

Try expanding first.

Answer 2

THE HINT SAID WE SHOULD EXPRESS b^2 in terms of a, and then solve for a....do not know what that means
\[b ^{2} = 4-a ^{2}\]

Answer 3

I just told you where to start.

Answer 4

ok, let me write what I found after expansion

Answer 5

\[a ^{3}+3a ^{2}ib-3ab ^{2}-ib ^{3}\]

Answer 6

Since the result is 8, the imaginary parts have to be equal to 0 because 8 is the same as 8+0i.

Answer 7

Carry on.

Answer 8

so
\[a ^{3}-3ab ^{2} =8\]
\[3a ^{2}b-b ^{3}=0\]

Answer 9

Yeah, and so on.

Answer 10

i want to assume that i should then solve for a^2 and b^2 and then substitute into \[a ^{2}-b ^{2}=4\] ?

Answer 11

Thought it was a +.

Answer 12

oh yes +

Answer 13

You can solve it however way you want.

Answer 14

thank u!

Answer 15

That method might be hard though. Just try manipulating the equations.

Answer 16

i can see that here, i am getting nowwhere

Answer 17

I'm actually getting 4/a, not just 4.

Answer 18

Ignore that.

Answer 19

it was an exam q , as is

Answer 20

So basically factor out the part that equals 0. You get that b=0 or 3a^2-b^2 = 0.
If b=0, then the other equation reduces to a^3 = 8. So a = 2 and a^2+b+2 = 4.
Now for the second case where 3a^2-b^2 = 0....

Answer 21

ok let me continue from there...

Answer 22

\[2^{2}-0^{2}=4\]

Answer 23

Eventually, you'll get another pair of solutions: a = -1 and b = sqrt3.

Answer 24

where is this one coming from?

Answer 25

There are always two solutions to a quad equation.