# Thread: [SOLVED] Multiplication with complex numbers

1. ## [SOLVED] Multiplication with complex numbers

I need to derive this equation and get the below result.

$\displaystyle ((a^2-b^2)^2+a^2b^2)(a^2+b^2)=a^6+b^6$

$\displaystyle ===>$ $\displaystyle (a^2-iab-b^2)(a+ib)=a^3-ib^3$

Also a hint teacher gave us:

replace

$\displaystyle a$ by $\displaystyle a^2$
$\displaystyle b$ by $\displaystyle -ib^2$

I tried few things (like $\displaystyle -b^2=-1.b^2=i^2b^2$) to involve complex numbers in the first equation but im confused and have no clue what to do.

2. Originally Posted by Nrt
I need to derive this equation and get the below result.

$\displaystyle ((a^2-b^2)^2+a^2b^2)(a^2+b^2)=a^6+b^6$

$\displaystyle ===>$ $\displaystyle (a^2-iab-b^2)(a+ib)=a^3-ib^3$

Also a hint teacher gave us:

replace

$\displaystyle a$ by $\displaystyle a^2$
$\displaystyle b$ by $\displaystyle -ib^2$

I tried few things (like $\displaystyle -b^2=-1.b^2=i^2b^2$) to involve complex numbers in the first equation but im confused and have no clue what to do.
Hi Nrt,

The first equation
$\displaystyle ((a^2-b^2)^2+a^2b^2)(a^2+b^2)=a^6+b^6$
is an identity-- it is true for all values of a and b.

Now
$\displaystyle (a^2-b^2)^2+a^2b^2 = a^4 - a^2b^2 +b^4$
so
$\displaystyle (a^4 - a^2b^2 +b^4)(a^2+b^2)=a^6+b^6$.

Now let $\displaystyle x = a^2$ and $\displaystyle y = b^2 / i$, so $\displaystyle b^2 = iy$.

Then the identity becomes
$\displaystyle (x^2 - x(iy) + (iy)^2)(x + iy) = x^3 + (iy)^3$
so
$\displaystyle (x^2 -i x y -y^2)(x+iy) = x^3 - i y^3$.

This is the result you were asked to show, except that I used x and y instead of a and b. It seems less confusing to me that way. (I get confused very easily). But you can go back and replace x and y with a and b if you prefer it that way.

3. Much appreciated, thank you.