4yz(z^(2)+6z-16)-:2y(z+8)

Simplify (4yz (z ^ (2) + 16z-16)) / (2y (z-8))

Factorise the expressions and divide them as directed. 4yz(z^2 + 6z - 16) div 2y(z + 8)

Simplify: 2-3z^(2)+5yz+7y^(2)-8+z^(2)-6yz-9y^(2)+1-2z^(2)-2yz-y^(2)

If x,y,z are real and 4x^(2) +9y^(2)+16z^(2) -6xy-12yz-8zx =0 then x,y,z are in

(y^(2)+yz+z^(2))/((x-y)(x-z))+(z^(2)+zx+x^(2))/((y-z)(y-x))+(x^(2)+xy+y^(2))/((z-x)(z-y))

(y^(2)+yz+z^(2))/((x-y)(x-z)) + (z^(2)+zx+x^(2))/((y-z)(y-x)) + (x^(2)+xy+y^(2))/((z-x)(z-y))

Find the product: (3x+2y+2z)(9x^(2)+4y^(2)+4z^(2)-6xy-4yz-6zx)

From the sum of 2y^(2)+3yz,-y^(2)-yz-z^(2) and yz+2z^(2) subtract the sum of 3y^(2)-z^(2) and -y^(2)+yz+z^(2)