Question

For all real numbers  x  and  y, if  x # y = x(x-y), then x # (x # y) =

My question is, how does x # (x # y) = 8 and how does it also equal to x(squared) - x(cubed) + x(squared) y?  

Is there a rule of the function that I have forgotten?  

Thanks

Answer 1

if x # y = x(x-y) => x # (x # y) = x # [x(x-y)] =x[ x - x(x-y)] = x( x - x^2 +x*y ) = x^2 - x^3 + ( x^2 ) * y;

Answer 2

`x\#y=x(x-y)\\\\x\#(x\#y)\\\\x\#y=x(x-y)=x^2-xy\\\\x\#(x\#y)=x\#(x^2-xy)=x[x-(x^2-xy)]=x(x-x^2+xy)\\\\=x^2-x^3+x^2y`



`x^2-x^3+x^2y=8\\\\x^2(1-x+y)=8\\\\x^2(1-x+y)=2^2\cdot4\iff x^2=2^2\ and\ 1-x+y=4\\\\x=2\ and\ y=4-1+x\\\\x=2\ and\ y=3+2\\\\x=2\ and\ y=5`