4/2 = d^3

solve for d - please show working

Question

4/2 = d^3

solve for d - please show working

anonymous · Answer

For this you have to ask your self what "undoes" the cube function

anonymous · Answer

ok, i understand if you take a square over it becomes a square root on other side right.

x^2 = 1
x=sqrt1

right?

anonymous · Answer

is it 

d = 3sqrt(4/2)

anonymous · Answer

Correct, you got it!

anonymous · Answer

thanks mate, its late and im doing a structural engineering subject, just wanted to check, all the best.

anonymous · Answer

one note: because a negative number times a negative number equals a positive, namely,
$$(-a)^2=a^2$$, you would have $$x^2=4 \implies x=+ or - 2$$
but this is not the case with odd numbered powers, i bet you could work this out too see why

anonymous · Answer

real and complex numbers is that right?

anonymous · Answer

no its not that complicated, it just has to do with what happens when you multiply a negaitive by a negative. $$-3*-3*-3 = -27, while 3*3*3 = 27, but -3*-3 = 9= 3*3$$