Melody

You cannot find the square root of a negative number 

So x is a positive square number.

and     \(0\le 63-3\sqrt{x} \le63\)

Now just test integer square values of x starting with x=0, then 1,4,9 etc

I think you have made this much more difficult and longer than necessary NotGuest.

You only need x^3 terms

These are the only terms that are relevant.      

\(6x^3*4=24x^3\\ 4x^2*-3x=-12x^3\\ 7x*3x^2=21x^3\\ -5*6x^3=-30x^3\\\)

Your answer makes no sense to me Hannah.  (MG)

Start by using the identity

sin(A-B)=sinAcosB-sinBcosA

This is almost identical to the question I just answered so you can do it yourself.

81.86%

^3√2x + 9^3√2x - 2^3√2x = 

Is 2x under the cube roo, of just the 2?  You should have used brackets.

Draw the triangle and use Pythagoras's theorem

Good  

angle CDA=angle CEB   corresponding angles on parrallel lines

If angle CAD = angle CBE what would have to be true about angle ACD and angle BCE?

I will let you finish it.