text: [(1/8)*x^2(2-x)^½]^2=?\n\nyou need these rules for calculations with Potens expression\n\na^n*b^n=(a*b)^n......@\na^n+a^m=a^(m+n)...@@\na^n^m=a^(n*m)........@@@\n(a/b)^n= a^n/b^n......@@@@\n\nNow to the work:\n\naccording to @\n[ (1/8)x^2 (2-x)^1/2 ]^2=\n((1/8)x^2)^2*((2-x)^½)^2=\n\naccording to @@@\n((1/8)x^2)^2*((2-x)^½)^2=\n(1/8)^2*x^4)*(2-x)^1)=\n(1/64)*x^4*(2-x)\n============\n\nyou see its have the roots x= 2 or x= 0\n==============================