Divide
`30x^(2)y^(3)z by -3xyz `

Divide -36x^(7)y^(6) z " by " - 12x^(2)y^(4)z^(3)

If x^(2)+y^(2)+z^(2)=1"for " x,y,z inR, then the maximum value of x^(3)+y^(3)+z^(3)-3xyz is

Divide: 25 x^3y^2\ by -15 x^2y (ii) -72 x^2y z\ by -12 x y z

Multiply : -3x^(2)y^(2) with -2xyz

Simplify (4x^(2)y^(2)z - 6xy^(3)z +10 x^(3)yz^(2)) div (-2xyz)

Factorise : 27x^(3)+8y^(3)+8z^(3)-36xyz

Verify that x^(3)+y^(3)+z^(3)-3xyz=(1)/(2)(x+y+z)[(x-y)^(2)+(y-z)^(2)+(z-x)^(2)]

Divide 24(x^2yz+xy^2z+xyz^2) by 8xyz using both the methods.

Using properties of determinants, prove that |{:(y + z ,z + x ,x + y ),(z + x ,x + y ,y + z),(x + y ,y + z,z + x ):}|=2 |{:(x, y, z),(y, z, x),(z, x, y):}|= - 2 (x^(3) + y^(3) + z^(3) - 3xyz)

Let x be the arithmetic mean and y, z be two geometric means between any two positive numbers. Then, prove that (y^(3) + z^(3))/(xyz) = 2 .